Sum of an Infinite Series

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jackiechad

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Apr 28, 2010
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Here is the problem I am working on:

The sum from 1 to infinity of (n^2)/(7^n)

I believe I need to somehow arrange this into a telescoping series but have had a tough time arranging it so that the terms work out for me. I would greatly appreciate any help you can give. Thanks so much!
 
Start with ∑k=1∞xk=x1−x,∣x∣<1\displaystyle \sum\limits_{k = 1}^\infty {x^k } = \frac{x}{{1 - x}},\quad \left| x \right| < 1.

Differenate and multiply x to get \(\displaystyle \[\sum\limits_{k = 1}^\infty {kx^k } = \frac{x}{{\left( {1 - x} \right)^2 }}\).

Do that again to get ∑k=1∞k2xk=x+x2(1−x)3\displaystyle \sum\limits_{k = 1}^\infty {k^2 x^k } = \frac{{x + x^2 }}{{\left( {1 - x} \right)^3 }}.

Then let x=17\displaystyle x=\frac{1}{7}.
 
Good show pka, I′m impressed.\displaystyle Good \ show \ pka, \ I'm \ impressed.
 
BigGlenntheHeavy said:
Good show pka, I′m impressed.\displaystyle Good \ show \ pka, \ I'm \ impressed.
Click to expand...
WHY? It is completely obvious to anyone having done similar problems.
 
Thank you SO much for the help, I was having an awful time with this problem. If you don't mind could you show the step where you differentiate, I'm a bit confused on that. Also, does plugging in 1/7 for x in the final equation give the sum? THANK YOU again!
 
jackiechad, here′s what pka did.\displaystyle jackiechad, \ here's \ what \ pka \ did.

Start with: ∑n=1∞xn =x1−x, ∣x∣ < 1\displaystyle Start \ with: \ \sum_{n=1}^{\infty}x^n \ =\frac{x}{1-x}, \ |x| \ < \ 1

Now, Dx[∑n=1∞xn =x1−x] = ∑n=1∞nxn−1 = 1(1−x)2\displaystyle Now, \ D_x\bigg[\sum_{n=1}^{\infty}x^n \ =\frac{x}{1-x}\bigg] \ = \ \sum_{n=1}^{\infty}nx^{n-1} \ = \ \frac{1}{(1-x)^2}

Then, x[∑n=1∞nxn−1 = 1(1−x)2] = ∑n=1∞nxn = x(1−x)2\displaystyle Then, \ x\bigg[ \sum_{n=1}^{\infty}nx^{n-1} \ = \ \frac{1}{(1-x)^2}\bigg] \ = \ \sum_{n=1}^{\infty}nx^n \ = \ \frac{x}{(1-x)^2}

Again, Dx[∑n=1∞nxn = x(1−x)2] = ∑n=1∞n2xn−1 = 1+x(1−x)3\displaystyle Again, \ D_x\bigg[\sum_{n=1}^{\infty}nx^n \ = \ \frac{x}{(1-x)^2}\bigg] \ = \ \sum_{n=1}^{\infty}n^2x^{n-1} \ = \ \frac{1+x}{(1-x)^3}

And finally: x[∑n=1∞n2xn−1 = 1+x(1−x)3] = ∑n=1∞n2xn = x+x2(1−x)3\displaystyle And \ finally: \ x\bigg[\sum_{n=1}^{\infty}n^2x^{n-1} \ = \ \frac{1+x}{(1-x)^3}\bigg] \ = \ \sum_{n=1}^{\infty}n^2 x^n \ = \ \frac{x+x^2}{(1-x)^3}

Now, letting x = 17, we get ∑n=1∞n27n = 17+149(67)3 = 727.\displaystyle Now, \ letting \ x \ = \ \frac{1}{7}, \ we \ get \ \sum_{n=1}^{\infty}\frac{n^2}{7^n} \ = \ \frac{\frac{1}{7}+\frac{1}{49}}{(\frac{6}{7})^3} \ = \ \frac{7}{27}.
 
Thank you for elaborating. This is a rather neat problem and certainly one that takes knowing that trick. Thank you both for the assistance
 
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