Thanks for watching! 😊 Check out ua-cam.com/video/pQ0dc8Oz4fE/v-deo.htmlsi=gmjJfkNN0EBC9zo0 for a really interesting infinite math sum and the 2 min simple creative trick to solve it!!! 🎉
Thanks Amitesh for being the one among the greatest to share what you have. You are the next Paul Erdos to me, to be the boy (of course a handsome man now) who loves Math and dedicate life for Math. Keep spreading the joy of learning Math!
Hi Nithya, thank you so much for such an incredibly kind and supportive comment! 😊 I am so happy to hear from you! 😊 I wish you a very happy New Year and all the absolute best to you and your family for 2025!!! 🥳🎉🎊
As someone who is just starting my journey in discrete math, this was a wonderful video. Your presentation style and excitement for the subject matter is great. I subscribed. I look forward to watching more videos! Thank you.
Hi @sirburr9220 thanks so much for your message! 😊 I really appreciate the positive feedback and I am so happy you loved the video! 🥳 I am excited to continue producing lots more math videos! 😊 I hope you have an amazing day/evening/night! 😊 Happy Holidays!!! 🥳🎉🎊
I was honestly a bit skeptical at first (I don't really know why though, it makes sense), but the way you explained it was very comprehensive! I'm subbing :)
Hi @justinferland6129 thank you so much for sharing your thoughts! 😊 I completely understand, it is indeed really surprising that 1 + 1/2 + 1/3 + 1/4 + ... should go on becoming larger and larger since eventually you are adding numbers like 1/10^1000 which are absolutely miniscule. The cool thing here is that it increases really, really slowly but still eventually does go above every number (exactly like log - if we take log to the base 10, then log_10(10^1000) = 1000 but 10^1000 is an absolutely enormous number, so it seems like log(x) will never reach infinity even as x -> infinity, but it does). The first time I saw this, I was skeptical too. 😅 I am so happy to hear that you subbed! 🥳 I hope you have an amazing day/evening/night! 😊
especially if you look at it as the infinite sum of 1/n^1. If we call this form 1/n^p, then in this case, p=1 (harmonic series). If p is greater than 1, even slightly, it does add up to a finite number. 1/n^1.0000000000001 adds up to a finite number. And your intuition makes perfect sense, each number is increasingly smaller, like Mr. Amitesh was explaining.
@@waspsandwich6548 Hi! Thank you so much for sharing that important perspective as well! It's definitely interesting to draw attention to the fact that the p-series sum_{n=1}^{∞} 1/n^p converges for each p > 1, although it diverges for p = 1. (The intuition for anyone else reading this is that as p -> 1 but p > 1, the sum_{n=1}^{∞} 1/n^p approaches infinity (as a function of p) although it is always finite as long as p > 1.) Thanks so much for sharing your insights and I hope you have an amazing day/evening/night! 😊
@@justinferland6129 Hi! I know this was a question addressed to someone else but for some reason I didn't get notified or see your comment here until now (just happened to notice it on my community tab). It is a great question! 🥳 The short answer is: this is what we call the "Riemann Zeta Function", one of the most important functions in mathematics. The value of the function at s is sum_{n=1}^{∞} 1/n^s (which makes sense for s > 1, although there is a process for defining the Riemann Zeta Function for all real values of s (that wouldn't be this sum in general since it diverges for s ≤ 1), and in fact, all complex number values of s). Anyway, I did a video on what sum_{n=1}^{∞} 1/n^2 is (you can check it out here - ua-cam.com/video/BbplJrw2gW8/v-deo.htmlsi=zpUmDkqTSvyZpwFh - it is π^2/6). I think it is quite a hard question to figure out in general. To this date, no-one knows what sum_{n=1}^{∞} 1/n^3 is (and even the case sum_{n=1}^{∞} 1/n^2 was a major open problem until Euler solved it in the 18th century). I hope you have an amazing day/evening/night! 😊
Hi @a_pc thank you so much for your extremely kind and supportive comment! 😊 I am so happy to read your comment and to know that you loved the proof. I hope you have an amazing day/evening/night! 😊
Damn Amit very good work! I was under the impression that this adds up to about 3.078 (something like that) but now I saw your explanation and it makes perfect sense. Very good work man I'm excited about what topics you will be covering in future. Subscribed! 🎉❤
Hi @thicccatto3956 thank you so much for your comment and for sharing your thoughts! 😊 I understand what you mean because it already takes roughly 16 terms of the series to exceed 3.078 (based on the bound in the video that adding up the first 2^n terms exceeds 1 + n/2), and if we looked at a small number like 10, it would take roughly a 1000 terms!! 🥳 I so happy to read your comment and to hear that you subscribed. I am also excited to make videos that you will enjoy watching in the future! 🥳 I hope you have an amazing day/evening/night! 😊
Hi @saravanabalajik thank you so much for your comment and for sharing your thoughts! 😊 I also love proofs that are as accessible as possible and proofs that reveal why something is true (more than just proving it is true), and this is why I made this video. 🥳 I hope you have an amazing day/evening/night! 😊
A fanstatic introduction on proving sequences by bounds, i encourage you to make more on things like sandwich theorem and even the epsilon N proofs, looking forward to your future videos
Hi @Aldehyde_ketone thanks so much for your kind and supportive comment, and amazing suggestions! 😊 I would definitely love to do more videos on those topics, I have already done a video on an epsilon-N proof (check out ua-cam.com/video/TLOueU-u87E/v-deo.htmlsi=Z3MO7_5f-Nkm76YP) and have a (growing) real analysis playlist which I started on fundamental theorems of sequences which I think will interest you (ua-cam.com/play/PL0NPansZqR_aO_1R6w50LfhKyEX1j4vuZ.html&si=Kg-RW3B93JVLMQHR). 🥳 I also have a video on the sandwich theorem in a fundamental example explained intuitively (ua-cam.com/video/pReC0squvck/v-deo.htmlsi=OjXa9PkhsMwsR3bx)! 🎁 In general, I have lots of content across all math topics and levels on my channel, so please feel free to check it out (especially the "Playlists" tab on my channel homepage to see stuff organized by topics and levels). However, I'm excited to continue creating videos that you will enjoy watching, and your comment gave me a great idea to do a rigorous proof of the sandwich theorem for limits! 😊 If you have thoughts for videos in general, please always feel free to let me know. Thanks for sharing your thoughts and for the support. I hope you have an amazing day/evening/night! 😊
I have a question on this. I like to think everything intuitively ( not just with the help of numbers) The sum of infinite series is s= a/(1-r) where r is the common ratio. For a infinite series like: 1/3+1/4+1/16+1/32 ..... This never reaches infinity. I always thought that yes, each time we are adding numbers but they are getting smaller, so at one point the change will be soo little that it won't matter. Same goes for the definition of e. e=limit x➡️♾️ (1+1/x)^x.. the more we calculate the value should get larger, but at a very low margin which is insignificant. So my question is where do we draw the line: if the growth rate is significant or not? In both cases the terms are getting smaller, but what is the fundamental difference between the 2 infinite series that makes one infinity and other a finite number? I hope you understand my question
Hi @rafiulhasan1055 thanks so much for your comment! 😊 Yes, this is a great question and the short answer is: it is a very deep question and there isn't a known "precise" boundary separating convergent series (where the sum is finite) and divergent series (where the sum is infinite). It is very interesting because (for example) the following cases are noted in Rudin's textbook "Principles of Mathematical Analysis" (1) sum_{n=1}^{∞} 1/n diverges (as in the video), but sum_{n=1}^{∞} 1/n^2 converges (will have a video on that soon). (2) sum_{n=2}^{∞} 1/(n log(n)) diverges, but sum_{n=2}^{∞} 1/(n (log(n))^2) converges (3) sum_{n=3}^{∞} 1/(n log(n) log(log(n))) diverges, but sum_{n=3}^{∞} 1/(n log(n) (log(log(n)))^2) converges so it seems like there should be a boundary, but there isn't a precise one (Rudin also says this in his textbook). If you saw my other videos on infinities recently, it's a very similar situation. I proved there that ∞ < 2^∞ and for a long time it was a question whether there is an "infinity strictly in between ∞ and 2^∞" and serious people (like Cantor himself) tried very hard to prove there wasn't. Ultimately, it was shown that it is impossible to prove this (within the standard axiomatic model of set theory known as ZF) and it became an independent hypothesis known as "the continuum hypothesis". I think. you can keep on finding examples of series that are very close to each other where one converges and the other diverges and there isn't any way to make a precise fine line distinction. Ultimately, we have to analyze each series on its own merits and there are various tests in real analysis that help to do so with many of the series that occur in "the wild". The geometric series you mentioned in your comment is an important example of a convergent series and many tests (like the ratio/root tests for convergence if you have seen them) are based on the fact that the terms in the geometric series become small "sufficiently fast" that the sum does converge. I hope that gives some insight into your question (not really an answer because I think your question is quite deep and no-one really knows how to precisely answer it/it may not even be possible to precisely answer it). I hope you have an amazing day/evening/night! 😊
Hi @saadiala-mb8zc thank you so much for sharing your amazing insights! 🥳 Yes, that is a great way of looking at it and also shows that the partial sums of the harmonic series have logarithmic growth! 😊 It's a beautiful observation based on the mean value theorem! Thanks again for sharing and I hope you have an amazing day/evening/night! 😊
A different proof is to use that sum (1/k ,k =1 to N) >∫1/x ,(1≤ x≤N)= ln N. The advantage is that one also can see how the sum diverges. Your proof has the advantage that one does not have know integration.
Hi @renesperb thank you so much for sharing this beautiful alternative approach and as always, for your comments and thoughts on the videos! 😊 I love your approach using the integral test for convergence/divergence and it deserves a video as well for calculus students! 🥳 As you say, the two methods have different pros/cons. I think the integral method is quicker once someone masters calculus. On the other hand, I think the method in the video also reveals how the sum diverges since it establishes an explicit bound sum_{k=1}^{2^n} 1/k ≥ 1 + n/2 and you can really see how the fractions are adding up as in the video (which may not be apparent from antidifferentiation). Of course, both methods are important to learn, and I am very happy you shared this integration method so people who know calculus can understand the harmonic series in another way by reading your comment. Thank you so much for sharing your thoughts and alternative method! 😊 I hope you have an amazing day/evening/night! 😊
By the way, I understand you're keep the production process as simple as possible but if you ever find a way to show the white board more clearly, be it close up pictures or whatever I think it would look even better. Just my 2 cents. Thanks again.
Hi @thiagof414 thank you so much for your feedback! I really appreciate it! 😊 You are probably right that it would help if the whiteboard took up more of the frame. The main thing I'm struggling with is making sure I also fit in the frame (especially when I'm on the extreme left/right). 😅 However, in recent videos I have recorded, I have tried to zoom in more/position the camera closer and I have been mindful of this (in this particular video, it could have been zoomed in more but I think I recorded it a long time ago (only uploaded recently)). I really appreciate your feedback and I hope you have an amazing day/evening/night! 😊
*Simple Proof using Elementary Integration of the simplest Geometric Series* Geometric series sum 1 + x + x^2 + x^3 + ... + x^n + ... = 1/(1-x) for x < 1 or tending to 1 from lower side. Integrate both sides to get -ln(1-x) or ln (1/(1-x)) = x + x^2/2 + x^3/3 + ... [no constant of integration because it is equal to -ln(1-0) = -ln(1) = 0 ] Therefore, at x = 1, we get 1 + 1/2 + 1/3 + ... + 1/n + ... = - ln (1-1) = ln (1/0) = ln (infinity) = infinity (hence a divergent series with positive infinite sum)
Hi @vishalmishra3046 thanks so much for sharing this beautiful proof method! 😊 Yes, it is another cool way of looking at it, and all the ideas are amazing, although some more justification around taking a (left-hand) limit as x -> 1 may be necessary to be completely rigorous (nonetheless, it does work). However, it is definitely a beautiful way of looking at it, and I think what is really cool about your method is that it applies to finding other interesting sums! For example, we can plug in values such as x = -1 to get the "alternating harmonic series" 1 - 1/2 + 1/3 - 1/4 + ... and find its sum is ln(2)! 🥳 Thanks so much for sharing and I hope you have an amazing day/evening/night! 😊
Thanks for watching! 😊 Check out ua-cam.com/video/pQ0dc8Oz4fE/v-deo.htmlsi=gmjJfkNN0EBC9zo0 for a really interesting infinite math sum and the 2 min simple creative trick to solve it!!! 🎉
Thanks Amitesh for being the one among the greatest to share what you have. You are the next Paul Erdos to me, to be the boy (of course a handsome man now) who loves Math and dedicate life for Math. Keep spreading the joy of learning Math!
Hi Nithya, thank you so much for such an incredibly kind and supportive comment! 😊 I am so happy to hear from you! 😊 I wish you a very happy New Year and all the absolute best to you and your family for 2025!!! 🥳🎉🎊
As someone who is just starting my journey in discrete math, this was a wonderful video. Your presentation style and excitement for the subject matter is great. I subscribed. I look forward to watching more videos! Thank you.
Hi @sirburr9220 thanks so much for your message! 😊 I really appreciate the positive feedback and I am so happy you loved the video! 🥳 I am excited to continue producing lots more math videos! 😊 I hope you have an amazing day/evening/night! 😊 Happy Holidays!!! 🥳🎉🎊
I was honestly a bit skeptical at first (I don't really know why though, it makes sense), but the way you explained it was very comprehensive! I'm subbing :)
Hi @justinferland6129 thank you so much for sharing your thoughts! 😊 I completely understand, it is indeed really surprising that 1 + 1/2 + 1/3 + 1/4 + ... should go on becoming larger and larger since eventually you are adding numbers like 1/10^1000 which are absolutely miniscule. The cool thing here is that it increases really, really slowly but still eventually does go above every number (exactly like log - if we take log to the base 10, then log_10(10^1000) = 1000 but 10^1000 is an absolutely enormous number, so it seems like log(x) will never reach infinity even as x -> infinity, but it does). The first time I saw this, I was skeptical too. 😅
I am so happy to hear that you subbed! 🥳 I hope you have an amazing day/evening/night! 😊
especially if you look at it as the infinite sum of 1/n^1. If we call this form 1/n^p, then in this case, p=1 (harmonic series). If p is greater than 1, even slightly, it does add up to a finite number. 1/n^1.0000000000001 adds up to a finite number. And your intuition makes perfect sense, each number is increasingly smaller, like Mr. Amitesh was explaining.
@@waspsandwich6548 Hi! Thank you so much for sharing that important perspective as well! It's definitely interesting to draw attention to the fact that the p-series sum_{n=1}^{∞} 1/n^p converges for each p > 1, although it diverges for p = 1. (The intuition for anyone else reading this is that as p -> 1 but p > 1, the sum_{n=1}^{∞} 1/n^p approaches infinity (as a function of p) although it is always finite as long as p > 1.)
Thanks so much for sharing your insights and I hope you have an amazing day/evening/night! 😊
@@waspsandwich6548 what would 1/n^1.0000000000001 converge to? Or if it's too big of a number, what about something like 1/n^2 or 10?
@@justinferland6129 Hi! I know this was a question addressed to someone else but for some reason I didn't get notified or see your comment here until now (just happened to notice it on my community tab). It is a great question! 🥳 The short answer is: this is what we call the "Riemann Zeta Function", one of the most important functions in mathematics. The value of the function at s is sum_{n=1}^{∞} 1/n^s (which makes sense for s > 1, although there is a process for defining the Riemann Zeta Function for all real values of s (that wouldn't be this sum in general since it diverges for s ≤ 1), and in fact, all complex number values of s).
Anyway, I did a video on what sum_{n=1}^{∞} 1/n^2 is (you can check it out here - ua-cam.com/video/BbplJrw2gW8/v-deo.htmlsi=zpUmDkqTSvyZpwFh - it is π^2/6). I think it is quite a hard question to figure out in general. To this date, no-one knows what sum_{n=1}^{∞} 1/n^3 is (and even the case sum_{n=1}^{∞} 1/n^2 was a major open problem until Euler solved it in the 18th century).
I hope you have an amazing day/evening/night! 😊
Loved it. Thank you!
Hi @thiagof414 thanks so much for your kind and supportive comment! 😊 I am so happy to read it! I hope you have an amazing day/evening/night! 😊
This is the best proof I've ever seen
Hi @a_pc thank you so much for your extremely kind and supportive comment! 😊 I am so happy to read your comment and to know that you loved the proof. I hope you have an amazing day/evening/night! 😊
That was an elegant one 👌
Hi @alex_be9698 thank you so much for your comment! I am so happy you liked the video! 😊
Damn Amit very good work! I was under the impression that this adds up to about 3.078 (something like that) but now I saw your explanation and it makes perfect sense. Very good work man I'm excited about what topics you will be covering in future. Subscribed! 🎉❤
Hi @thicccatto3956 thank you so much for your comment and for sharing your thoughts! 😊 I understand what you mean because it already takes roughly 16 terms of the series to exceed 3.078 (based on the bound in the video that adding up the first 2^n terms exceeds 1 + n/2), and if we looked at a small number like 10, it would take roughly a 1000 terms!! 🥳
I so happy to read your comment and to hear that you subscribed. I am also excited to make videos that you will enjoy watching in the future! 🥳 I hope you have an amazing day/evening/night! 😊
Awesome, don't need integration, such thing in mathematics give me hope about our universe.
Hi @saravanabalajik thank you so much for your comment and for sharing your thoughts! 😊 I also love proofs that are as accessible as possible and proofs that reveal why something is true (more than just proving it is true), and this is why I made this video. 🥳 I hope you have an amazing day/evening/night! 😊
A fanstatic introduction on proving sequences by bounds, i encourage you to make more on things like sandwich theorem and even the epsilon N proofs, looking forward to your future videos
Hi @Aldehyde_ketone thanks so much for your kind and supportive comment, and amazing suggestions! 😊 I would definitely love to do more videos on those topics, I have already done a video on an epsilon-N proof (check out ua-cam.com/video/TLOueU-u87E/v-deo.htmlsi=Z3MO7_5f-Nkm76YP) and have a (growing) real analysis playlist which I started on fundamental theorems of sequences which I think will interest you (ua-cam.com/play/PL0NPansZqR_aO_1R6w50LfhKyEX1j4vuZ.html&si=Kg-RW3B93JVLMQHR). 🥳 I also have a video on the sandwich theorem in a fundamental example explained intuitively (ua-cam.com/video/pReC0squvck/v-deo.htmlsi=OjXa9PkhsMwsR3bx)! 🎁
In general, I have lots of content across all math topics and levels on my channel, so please feel free to check it out (especially the "Playlists" tab on my channel homepage to see stuff organized by topics and levels). However, I'm excited to continue creating videos that you will enjoy watching, and your comment gave me a great idea to do a rigorous proof of the sandwich theorem for limits! 😊 If you have thoughts for videos in general, please always feel free to let me know. Thanks for sharing your thoughts and for the support. I hope you have an amazing day/evening/night! 😊
I have a question on this. I like to think everything intuitively ( not just with the help of numbers)
The sum of infinite series is s= a/(1-r) where r is the common ratio. For a infinite series like: 1/3+1/4+1/16+1/32
..... This never reaches infinity. I always thought that yes, each time we are adding numbers but they are getting smaller, so at one point the change will be soo little that it won't matter. Same goes for the definition of e. e=limit x➡️♾️ (1+1/x)^x.. the more we calculate the value should get larger, but at a very low margin which is insignificant.
So my question is where do we draw the line: if the growth rate is significant or not? In both cases the terms are getting smaller, but what is the fundamental difference between the 2 infinite series that makes one infinity and other a finite number?
I hope you understand my question
Hi @rafiulhasan1055 thanks so much for your comment! 😊 Yes, this is a great question and the short answer is: it is a very deep question and there isn't a known "precise" boundary separating convergent series (where the sum is finite) and divergent series (where the sum is infinite). It is very interesting because (for example) the following cases are noted in Rudin's textbook "Principles of Mathematical Analysis"
(1) sum_{n=1}^{∞} 1/n diverges (as in the video), but sum_{n=1}^{∞} 1/n^2 converges (will have a video on that soon).
(2) sum_{n=2}^{∞} 1/(n log(n)) diverges, but sum_{n=2}^{∞} 1/(n (log(n))^2) converges
(3) sum_{n=3}^{∞} 1/(n log(n) log(log(n))) diverges, but sum_{n=3}^{∞} 1/(n log(n) (log(log(n)))^2) converges
so it seems like there should be a boundary, but there isn't a precise one (Rudin also says this in his textbook). If you saw my other videos on infinities recently, it's a very similar situation. I proved there that ∞ < 2^∞ and for a long time it was a question whether there is an "infinity strictly in between ∞ and 2^∞" and serious people (like Cantor himself) tried very hard to prove there wasn't. Ultimately, it was shown that it is impossible to prove this (within the standard axiomatic model of set theory known as ZF) and it became an independent hypothesis known as "the continuum hypothesis".
I think. you can keep on finding examples of series that are very close to each other where one converges and the other diverges and there isn't any way to make a precise fine line distinction. Ultimately, we have to analyze each series on its own merits and there are various tests in real analysis that help to do so with many of the series that occur in "the wild". The geometric series you mentioned in your comment is an important example of a convergent series and many tests (like the ratio/root tests for convergence if you have seen them) are based on the fact that the terms in the geometric series become small "sufficiently fast" that the sum does converge.
I hope that gives some insight into your question (not really an answer because I think your question is quite deep and no-one really knows how to precisely answer it/it may not even be possible to precisely answer it). I hope you have an amazing day/evening/night! 😊
Using the mean value theorem for ln(t) in some interval (k,k+1) where k is not zero positive integer
ln(k+1)-ln(k)
Hi @saadiala-mb8zc thank you so much for sharing your amazing insights! 🥳 Yes, that is a great way of looking at it and also shows that the partial sums of the harmonic series have logarithmic growth! 😊 It's a beautiful observation based on the mean value theorem! Thanks again for sharing and I hope you have an amazing day/evening/night! 😊
A different proof is to use that sum (1/k ,k =1 to N) >∫1/x ,(1≤ x≤N)= ln N. The advantage is that one also can see how the sum diverges. Your proof has the advantage that one does not have know integration.
Hi @renesperb thank you so much for sharing this beautiful alternative approach and as always, for your comments and thoughts on the videos! 😊 I love your approach using the integral test for convergence/divergence and it deserves a video as well for calculus students! 🥳 As you say, the two methods have different pros/cons. I think the integral method is quicker once someone masters calculus. On the other hand, I think the method in the video also reveals how the sum diverges since it establishes an explicit bound sum_{k=1}^{2^n} 1/k ≥ 1 + n/2 and you can really see how the fractions are adding up as in the video (which may not be apparent from antidifferentiation). Of course, both methods are important to learn, and I am very happy you shared this integration method so people who know calculus can understand the harmonic series in another way by reading your comment. Thank you so much for sharing your thoughts and alternative method! 😊 I hope you have an amazing day/evening/night! 😊
By the way, I understand you're keep the production process as simple as possible but if you ever find a way to show the white board more clearly, be it close up pictures or whatever I think it would look even better. Just my 2 cents. Thanks again.
Hi @thiagof414 thank you so much for your feedback! I really appreciate it! 😊 You are probably right that it would help if the whiteboard took up more of the frame. The main thing I'm struggling with is making sure I also fit in the frame (especially when I'm on the extreme left/right). 😅 However, in recent videos I have recorded, I have tried to zoom in more/position the camera closer and I have been mindful of this (in this particular video, it could have been zoomed in more but I think I recorded it a long time ago (only uploaded recently)). I really appreciate your feedback and I hope you have an amazing day/evening/night! 😊
*Simple Proof using Elementary Integration of the simplest Geometric Series*
Geometric series sum 1 + x + x^2 + x^3 + ... + x^n + ... = 1/(1-x) for x < 1 or tending to 1 from lower side.
Integrate both sides to get -ln(1-x) or ln (1/(1-x)) = x + x^2/2 + x^3/3 + ... [no constant of integration because it is equal to -ln(1-0) = -ln(1) = 0 ]
Therefore, at x = 1, we get 1 + 1/2 + 1/3 + ... + 1/n + ... = - ln (1-1) = ln (1/0) = ln (infinity) = infinity (hence a divergent series with positive infinite sum)
Hi @vishalmishra3046 thanks so much for sharing this beautiful proof method! 😊 Yes, it is another cool way of looking at it, and all the ideas are amazing, although some more justification around taking a (left-hand) limit as x -> 1 may be necessary to be completely rigorous (nonetheless, it does work). However, it is definitely a beautiful way of looking at it, and I think what is really cool about your method is that it applies to finding other interesting sums! For example, we can plug in values such as x = -1 to get the "alternating harmonic series" 1 - 1/2 + 1/3 - 1/4 + ... and find its sum is ln(2)! 🥳 Thanks so much for sharing and I hope you have an amazing day/evening/night! 😊