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Let $x\in(0,1)$. My goal is to compute this well known almost-geometric series. $$\sum_{k\geq0}(k+1)x^{k}$$

I've computed it using two similar methods, leading to different results. That pineapple brain of mine can't find the error.

method 1 :

\begin{align} \sum_{k\geq0}(k+1)x^{k}=& \sum_{k\geq-1}(k+1)x^{k} && \text{since }k=-1\Rightarrow (-1+1)x^{-1}=0 \\ =&\sum_{\tilde{k}\geq0}\tilde{k}\cdot x^{\tilde{k}-1} && \text{letting } \tilde{k}-1=k\\ =&\sum_{\tilde{k}\geq0}\frac{d}{dx}x^{\tilde{k}}\\ =&\frac{d}{dx}\sum_{\tilde{k}\geq0}x^{\tilde{k}} \end{align}

which is the derivative of an easily-computable geometric series.

method 2 :

\begin{align} \sum_{k\geq0}(k+1)x^{k}=& 1 + \sum_{k\geq1}(k+1)x^{k} && \text{since }k=0\Rightarrow (0+1)x^0=1 \\ =& 1 + \frac{d}{dx}\sum_{k\geq1}x^{k+1}\\ =& 1 + \frac{d}{dx}\sum_{k\geq1}x^{k+1} \\ =& 1 + \frac{d}{dx}\sum_{\tilde{k}\geq0}x^{\tilde{k}} && \text{letting } \tilde{k}-1=k\\ \end{align}

And a $1$ has been created.

Any help would be greatly appreciated.

Bernard
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lou
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    $\displaystyle\sum_{k\ge 1}x^{k+1}=\sum_{\bar k\ge 0} x^{\bar k+2}\ne\sum_{\bar k\ge 0} x^{\bar k}$, the last step. – Prasun Biswas Aug 01 '21 at 18:26
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    In your last formula $\tilde k$ should start at $2$ and not $0$. – Gary Aug 01 '21 at 18:27
  • Thank you, greatly appreciated. – lou Aug 01 '21 at 18:28
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    This seems overly complicated. Why don't you simply write that $;\sum\limits_{k\geq0}(k+1)x^{k}=\dfrac{\mathrm d}{\mathrm dx}\sum\limits_{k\geq 0}x^{k}$? – Bernard Aug 01 '21 at 18:31
  • To rigorously change the derivative and the series, see https://math.stackexchange.com/a/147882/631742 – Maximilian Janisch Aug 01 '21 at 18:37
  • @Bernard Simply because $\frac{d}{dx}\sum_{k\geq0}x^k = \sum_{k\geq0}k\cdot x^{(k-1)}$ wich isn't immediatly trivial to me that it is indeed $\sum_{(k\geq0)}(k+1)x^k$. Thanks for your concern and your advice. – lou Aug 01 '21 at 18:41
  • Thank you @MaximilianJanisch – lou Aug 01 '21 at 18:42

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