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?
2026-03-28 11:13:36.1774696416
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Sum of this series $\sum\limits_{n=1}^{\infty}e^nx^ne^{-xn}$?
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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*}
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, $x < e^{x-1}$