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What is the sum of this series? $$\sum_{n=1}^{\infty}e^nx^ne^{-xn}$$ I tried using the geometric series sum on $$\sum_{n=1}^{\infty}\left(\dfrac{ex}{e^x}\right)^n$$ and got $\dfrac{1}{1-\dfrac{xe}{e^x}}$ but i'm not sure if this is right?

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    $\begingroup$ Your result is fine, subject to the convergence condition that $|{xe\over e^x}|\lt1$. $\endgroup$ – Barry Cipra Apr 26 '17 at 14:37
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    $\begingroup$ And you must add a multiplicative factor $ex/\exp(x)$, as the summation begins at $n=1$. $\endgroup$ – Kelenner Apr 26 '17 at 14:40
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You missed to multiply the first term, i.e. the result will be :

$$\dfrac{\dfrac{xe}{e^x}}{1-\dfrac{xe}{e^x}}$$

provided $$\Bigg|\dfrac{xe}{e^x} \Bigg| <1$$

i.e. $$\forall x \in \mathbb R - \{1\} $$

Since at $x=1$, the series becomes $1+1+1+1 \ldots$ which obviously diverges. Everywhere else, $x < e^{x-1}$

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  • $\begingroup$ can you tell me how his result is fine? as $n$ starts from 1, so there must be a multiplicative factor $\dfrac{ex}{e^x}$ with his ans. $\endgroup$ – k.Vijay Apr 26 '17 at 15:02
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    $\begingroup$ @k.Vijay Really sorry, I just went with the flow andd didn't notice that the sum starts from $n=1$. Now I've corrected. $\endgroup$ – Jaideep Khare Apr 26 '17 at 15:20
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No, your ans is not right, as $n$ starts from $1$. Follow this: \begin{align*} \sum\limits_{n=1}^{\infty}e^nx^ne^{-xn}&=\sum\limits_{n=1}^{\infty}\left(\dfrac{ex}{e^x}\right)^n\\ &=\dfrac{\dfrac{ex}{e^x}}{1-\dfrac{ex}{e^x}}\hspace{30pt}\text{ as, }\left|\dfrac{ex}{e^x} \right| <1\\ &=\dfrac{1}{\dfrac{e^x}{ex}-1}\\ &=\dfrac{ex}{e^x-ex} \end{align*}

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