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?

  • 3
    $\begingroup$ Your result is fine, subject to the convergence condition that $|{xe\over e^x}|\lt1$. $\endgroup$ – Barry Cipra Apr 26 '17 at 14:37
  • 4
    $\begingroup$ 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

You missed to multiply the first term, i.e. the result will be :


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, $x < e^{x-1}$

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.