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So I am studying series for an exam right now and there is an example in the book I am studying (unfortunately the book is specific to my university so I cannot give any link) where certain series' sums are calculated. At some point the following ones are calculated:

$$\sum_{n=1}^{\infty}\left(\dfrac{1}{5}\right)^{n-1}=\frac54$$

and $$\sum_{n=1}^{\infty} n\left(\dfrac{1}{5}\right)^{n-1}=\frac{25}{16}$$

I do not understand the reason that extra $n$ (the multiplier in the front) squares everything. Also I have no clue what to do when I have say

$$\sum_{n=1}^{\infty}n^2\left(\dfrac{1}{5}\right)^{n-1}$$

Do I cube everything? Is there a specific way to do it that I do not get? If there is some online paper, book chapter or whatever that could help me, please link me to it!

Angelos Chalaris
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    What you are looking for is derivation of series of the form $\sum_{n=0}^\infty x^n = 1/(1-x)$. Take the derivative you'll get the extra $n$ you mention. Take the second derivative you get $n(n-1)$ (and not $n^2$ but you can find the result with the two previous ones). You need some theory to do that (radius of convergence etc...) but that's roughly the idea – Bertrand R Aug 12 '14 at 11:14
  • @BertrandR any specific page or paper with examples you could direct me to? I know some theory is missing from the book and probably this part as well so I could use some other source to get a good grasp of this! – Angelos Chalaris Aug 12 '14 at 11:24
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    Did you try the WP page on "geometric series"? – Did Aug 12 '14 at 11:35
  • Google 'Cauchy product'. When you square the first sum, the n is left from the simplification of the inner sum. – Caddyshack Aug 12 '14 at 13:16

1 Answers1

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if $|x|<1$ then Geometric series says: \begin{equation}\sum_{k=0}^{\infty}x^k=\dfrac{1}{1-x}\qquad (1)\end{equation} In your case substitute $n-1=s$ and then use it.

Derivating $(1)$ we get \begin{align*}\sum_{k=0}^{\infty}kx^{k-1}&=\dfrac{1}{(1-x)^2}\\ \iff\sum_{k=1}^{\infty}kx^{k-1}&=\dfrac{1}{(1-x)^2} \end{align*}

Now multiply by $x$ $$\sum_{k=1}^{\infty}kx^{k}=\dfrac{x}{(1-x)^2}$$

Derivate again

$$\sum_{k=1}^{\infty}k^2x^{k-1}=-\dfrac{x+1}{(x-1)^3}$$

Bman72
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  • This is exactly what I was looking for! Thanks a lot, you saved me a lot of headaches! :D – Angelos Chalaris Aug 12 '14 at 11:46
  • @Spiritios No problem :P remember the first formula that if $|x|<1$ then $\sum_k x^k=\frac{1}{1-x}$ the other can always be derivated from this one and is also a good exercise :D – Bman72 Aug 12 '14 at 11:48