I actually started with S(x) = SUM(j=0..inf; (2*j+1)*x^(2*j)), and evaluated the generating function this gave me at x = 1/2. One thing I want to point out - @7:24, you can multiply 1/(1 - x) by itself. This is more laborious than taking the derivative, but you end up in the same place, without having to take a detour through differential calculus.
Ah, but some just LOVE calculus, me too... I also started with S(x) = SUM(j=0..inf; (2*j+1)*x^(2*j)), which obviously is the derivative(!) of SUM(j=0..inf; x^(2*j+1)), which is equal to x*SUM(j=0..inf; (x^2)^j), which is equal to x/(1-x^2). So take derivative of that, which is (1+x^2)/(1-x^2)^2 and evaluate at 1/2...
I found your channel recently. Love your videos. Keep it up.
I appreciate it! 😄
the ratio test shows this series is absolutely convergent
I joined half way through the premiere - at Method 1 - enjoyed that
Glad you enjoyed it!
I actually started with S(x) = SUM(j=0..inf; (2*j+1)*x^(2*j)), and evaluated the generating function this gave me at x = 1/2. One thing I want to point out - @7:24, you can multiply 1/(1 - x) by itself. This is more laborious than taking the derivative, but you end up in the same place, without having to take a detour through differential calculus.
Ah, but some just LOVE calculus, me too...
I also started with S(x) = SUM(j=0..inf; (2*j+1)*x^(2*j)),
which obviously is the derivative(!) of SUM(j=0..inf; x^(2*j+1)),
which is equal to x*SUM(j=0..inf; (x^2)^j),
which is equal to x/(1-x^2).
So take derivative of that, which is (1+x^2)/(1-x^2)^2 and evaluate at 1/2...
5:22 Siri in my Iphone responded for you here🤣
😁😍
1/1 + 4/4 + 9/16 + 16/64 + 25/256 + ... = ?
Hint: the result from this video can be used to sum the above series.
I used a different method:
Let f(x) = 1 + 2x + 3x³ + ... = 1/(1-x)²
Series = ½(f(½) + f(-½)) = ½(4 + 4/9) = 20/9.
Nice approach!
Again AGP
S= 1+ 3/4 + 5/16 + 7/64 + 9/256 + 11/1024+..... (Each term is being multiplied by1/4)
S/4= 1/4 + 3/16 + 5/64 + 7/256 + 9/1024 [Subtracting them]
S-S/4= 1 + 2/4 + 2/16 + 2/64 + 2/256 + 2/1024+....
3S/4=1+2(1/4+1/16+1/64+1/256+1/1024+......)
Now the second part forms a infinite geometric series,
Sum of infinite geometric series= a/(1-r) [a= first term; r= ratio]
3S/4=1+2[(1/4)/(1-1/4)]=1+2[(1/4)/(3/4)]=1+2/3
S=(4/3)*(5/3)
S=20/9
Nice - I almost got the correct answer.
That's great! Do you see where you made the mistake?
@@SyberMath no - I only took a few partial sums and stopped there.
Thank you for saying ramanujan sum is false
THANK YOU
No problem! I needed to say that
I got 20/9 after 50 terms lets see if i was right
Σ(2k+1)/4^k=2*(1/4)/(1-1/4)^2+1/(1-1/4)=8/9+4/3=20/9
Σ (2n + 1) / 4^n = Σ (2n + 1) x^n for x = 1/4 and n = 0, 1, 2, 3, ...
Σ (2n + 1) x^n = Σ (-1 + 2 + 2x + 2x² + 2x³ + ...) x^n = (-1 + 2 + 2x + 2x² + 2x³ + ...) 1 / (1 - x)
=[ -1 + 2 / (1 - x) ] / (1 - x) = (1 + x) / (1 - x)² = 20/9 for x = 1/4
Haven't we just done this one? A repetitive pattern of videos.