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I am trying to prove how $$g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}$$

or $\sum ak(k-1)r^(k-1) = 2a(1-r)^{-3}$.

I don't know what I am doing wrong and am at my wits end.

The attempt at a solution (The index of the summation is always $k=2$ to infinity)

\begin{align*} \sum ak(k-1)r^{k-1} &=a \sum k(k-1)r^{k-2}\\ &=a \sum (r^k)''\\ &=a \left(\frac{r^2}{1-r}\right)'' \end{align*}

From this point I get a mess, and the incorrect answer. The thing I have a problem is I think $\sum (r^k)''$ when $k$ is from $2$ to infinity is $r^2/(1-r)$, since the first term in this sequence is $r^2$. I don't think that is correct though.

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1 Answer 1

$$f(x)=\sum_{k=0}^\infty ax^k=\frac{a}{1-x}\;\;,\;\;|x|<1$$

$$f'(x)=\sum_{k=1}^\infty akx^{k-1}=\frac{a}{(1-x)^2}\;\;,\;\;|x|<1$$

$$f''(x)=\sum_{k=2}^\infty ak(k-1)x^{k-2}=\frac{2a}{(1-x)^3}\;\;,\;\;|x|<1$$

Withing the convergence interval, you can derivate/integrate elementwise.

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I'm sorry but I don't understand what you asked. –  DonAntonio Nov 8 '12 at 14:31
sorry, i removed the comment, and would remove this question as well after I realised how stupid I had been –  user929404 Nov 8 '12 at 16:06

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