We know that, for -1 < x < 1: 1/(1-x) = 1 + x + x^2 + x^3 + … Taking the derivative of both sides: 1/(1-x)^2 = 1 + 2x + 3x^2 + 4x^3 + … Now multiply both sides by x: x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 + … Plugging in x=5 gives us the series that we want: 1/5 + 2/25 + 3/125 + … = (1/5)/(1-1/5)^2 = 1/5 * (5/4)^2 = 1/5 * 25/16 = 5/16
Very cool. In this example, a=r=⅕ without the arithmetic progression, so the nth term is ar^(n-1)=rⁿ or ⅕ⁿ, but adding the arithmetic bit, we get aₙ=nr and an nth term of nrⁿ. Next, you show that S-rS=a/(1-r), thus S=a/(1-r)². But a=r, so S=r/(1-r)². So changing from a constant "a" to an arithmetic aₙ changed the sum by a factor of 1/(1-r) when a=r and aₙ=nr. Letting r=1/k, then the sum is k/(k-1) times bigger with the arithmetic aₙ. So when k=2, or r=½, it exactly doubles.
Another method to use calculus Sum from 1 to inf of x^n = x/(1-x) Differentiate both side: Sum from 1 to inf of nx^(n-1) = 1/(1-x)^2 Multply both side by x: Sum from 1 to inf of nx^n = x/(1-x)^2 Set x=1/5 We get the sum equal to 5/16
This relies on these series being absolutely convergent so that the order of the sums can be manipulated with regards to associativity and commutativity when you add the infinite series. But being more rigorous doesn’t subtract from ingenuity of the solution. Cool!
I solved this by setting the sum equal to S, and noting that S - S/5 is the geometric series with common ratio 1/5. And that got me interested in the following sum: SUM(j=1..inf; j^2/5^j). Finding a value for this one just requires knowing that (n+1)^2 - n^2 = 2*n+1, and the value of the sum in this video.
Couldn’t do this problem for the longest time maybe Weeks and then yesterday as I was falling asleep in my rocking chair only half awake the idea to multiply the series by five and subtract The series appeared from somewhere in my unconscious. Strange and fascinating to me how the mind works when you’re not paying attention.
these are 2 infinite sums. Generally speaking they cannot be subtracted. How are we sure that the subtraction is valid and the result is not something like the -1/12
Because the sum converges, when the sum diverges you cannot do it without setting an axiom that allows you to. (Similar to the fact that (i)^2=-1 does not really exist so we celebrated ab axiom for it for practical uses)
Alternative solution. From formula at 1:40, we see that Σ i=0 to ∞ ((1/5)^i) = 1/(1-1/5) = 1/(4/5) = 5/4. Thus Σ i=1 to ∞ ((1/5)^i) = (5/4)-1 = 1/4 - it is the same as the previous sum, less the first element when i=0, the value of the first element being (1/5)^0 = 1. Back to our challenge at hand. Σ i=1 to ∞ (i/((1/5)^i)) = Σ j=1 to ∞ ( Σ i=j to ∞ (1/((1/5)^i)) ) = Σ j=1 to ∞ ( 1/((1/5)^(j-1)) * Σ i=1 to ∞ (1/((1/5)^i)) ) = Σ j= 1 to ∞ ( 1/((1/5)^(j-1))) * 1/4) = 1/4 * Σ j= 0 to ∞ ( 1/((1/5)^j)) = 1/4 * 5/4 = 5/16.
Overdoing it as a little practice: Summation from n=1 to n tends to inf, of n/x^n, where value at x = 5 Rewrite it: summation of n*(x^-n) pull out a (-x), i.e. (-x) * summation of (-n)*(x^(-n-1)) = (-x) * (summation of d/dx x^-n) = (-x) * d/dx (summation of x^-n), change the order legitimately. = (-x) * d/dx 1/(1-x^-1) , note that common ratio is x^-1, which is ultimately 1/5 less than 1. = (-x) * d/dx x/(x-1) = (-x) * (-1/(x-1)^2) = x/(x-1)^2 required ans: plug x=5, 5/16 follows.
We know that, for -1 < x < 1:
1/(1-x) = 1 + x + x^2 + x^3 + …
Taking the derivative of both sides:
1/(1-x)^2 = 1 + 2x + 3x^2 + 4x^3 + …
Now multiply both sides by x:
x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 + …
Plugging in x=5 gives us the series that we want:
1/5 + 2/25 + 3/125 + …
= (1/5)/(1-1/5)^2
= 1/5 * (5/4)^2
= 1/5 * 25/16 = 5/16
Generating function - lethal weapon indeed
@@vladimirkaplun5774Really just a power series
Great stuff!
@@seanfraser3125 It is exactly Generating Function method.
Very nice!
Very cool. In this example, a=r=⅕ without the arithmetic progression, so the nth term is ar^(n-1)=rⁿ or ⅕ⁿ, but adding the arithmetic bit, we get aₙ=nr and an nth term of nrⁿ. Next, you show that S-rS=a/(1-r), thus S=a/(1-r)². But a=r, so S=r/(1-r)². So changing from a constant "a" to an arithmetic aₙ changed the sum by a factor of 1/(1-r) when a=r and aₙ=nr. Letting r=1/k, then the sum is k/(k-1) times bigger with the arithmetic aₙ. So when k=2, or r=½, it exactly doubles.
Another method to use calculus
Sum from 1 to inf of x^n = x/(1-x)
Differentiate both side:
Sum from 1 to inf of nx^(n-1) = 1/(1-x)^2
Multply both side by x:
Sum from 1 to inf of nx^n = x/(1-x)^2
Set x=1/5
We get the sum equal to 5/16
Very interesting, thank you so much!
My pleasure!
That was amazing!
This relies on these series being absolutely convergent so that the order of the sums can be manipulated with regards to associativity and commutativity when you add the infinite series. But being more rigorous doesn’t subtract from ingenuity of the solution. Cool!
Cool solution👍
Can be generalized to k/(k-1)^2 for any infinite sum n/(k^n) where n goes from 1 to infinity assuming the sum converges.
Beautiful question...❤❤❤.
Learn alot...
Glad it was helpful!
I solved this by setting the sum equal to S, and noting that S - S/5 is the geometric series with common ratio 1/5. And that got me interested in the following sum: SUM(j=1..inf; j^2/5^j). Finding a value for this one just requires knowing that (n+1)^2 - n^2 = 2*n+1, and the value of the sum in this video.
Couldn’t do this problem for the longest time maybe Weeks and then yesterday as I was falling asleep in my rocking chair only half awake the idea to multiply the series by five and subtract The series appeared from somewhere in my unconscious. Strange and fascinating to me how the mind works when you’re not paying attention.
Exactly! Remember how Benzen’s chemical structure was discovered by Kekule.
these are 2 infinite sums. Generally speaking they cannot be subtracted.
How are we sure that the subtraction is valid and the result is not something like the -1/12
Because the sum converges, when the sum diverges you cannot do it without setting an axiom that allows you to. (Similar to the fact that (i)^2=-1 does not really exist so we celebrated ab axiom for it for practical uses)
1/5(1/(1-1/5)^2)=5/16...derivata della serie geometrica
Alternative solution.
From formula at 1:40, we see that Σ i=0 to ∞ ((1/5)^i) = 1/(1-1/5) = 1/(4/5) = 5/4.
Thus Σ i=1 to ∞ ((1/5)^i) = (5/4)-1 = 1/4 - it is the same as the previous sum, less the first element when i=0, the value of the first element being (1/5)^0 = 1.
Back to our challenge at hand.
Σ i=1 to ∞ (i/((1/5)^i)) =
Σ j=1 to ∞ ( Σ i=j to ∞ (1/((1/5)^i)) ) =
Σ j=1 to ∞ ( 1/((1/5)^(j-1)) * Σ i=1 to ∞ (1/((1/5)^i)) ) =
Σ j= 1 to ∞ ( 1/((1/5)^(j-1))) * 1/4) =
1/4 * Σ j= 0 to ∞ ( 1/((1/5)^j)) =
1/4 * 5/4 = 5/16.
No worries. Here is sigma: Σ and infinity: ∞
@@SyberMath Thanks. I edited it, and it does look elegant!
@@draganminic4928 you rock! 🤩
Overdoing it as a little practice:
Summation from n=1 to n tends to inf, of n/x^n, where value at x = 5
Rewrite it: summation of n*(x^-n)
pull out a (-x), i.e. (-x) * summation of (-n)*(x^(-n-1))
= (-x) * (summation of d/dx x^-n)
= (-x) * d/dx (summation of x^-n), change the order legitimately.
= (-x) * d/dx 1/(1-x^-1) , note that common ratio is x^-1, which is ultimately 1/5 less than 1.
= (-x) * d/dx x/(x-1)
= (-x) * (-1/(x-1)^2)
= x/(x-1)^2
required ans: plug x=5, 5/16 follows.
Similarly, x D_x 1/(1- x) = x/(1 - x)^2 evaluated at x = 1/5.
Here denominator is in GP and numerator in AP we called this as Arithematic Geometric Progression , Sir I am an IIT JEE Aspirant.
Solved in 1st attempt by your method only
5/16