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I need to calculate $\sum_{n=1}^\infty \dfrac{nx}{e \ ^ {nx}}$ for $x \ge0 $.

Thanks !

  • $k$ doesn't appear in your sum. Either change it to $n$ or change your $n$'s to $k$'s. – JasonM Jul 27 '16 at 06:54
  • If your series is accurate then it diverges big time: either those $;x'$ s are $;k'$ s, or else change the $;k;$ in the series lower limit for an $;x;$ . – DonAntonio Jul 27 '16 at 06:55
  • In your integrals, are you integrating with respect to $n$ or $x$? – Henry Jul 27 '16 at 06:55
  • @JasonM changed it, sorry for that –  Jul 27 '16 at 06:56
  • @Henry integral with respect o x –  Jul 27 '16 at 06:56
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    This is $x\sum_{n=1}^\infty nr^n$ with $r=e^{-x}$, and you may want to check out http://math.stackexchange.com/questions/30732. – Jonas Meyer Jul 27 '16 at 06:57
  • But in your sum, $x$ is exogenous while $n$ goes from $1$ to $\infty$ – Henry Jul 27 '16 at 06:57
  • @Liad No worries. Don't forget to put the $dx$ or whichever you meant in your integrals. And you can't equate the sum and integral. It's just bounded by it – JasonM Jul 27 '16 at 06:58
  • @Henry Well you are right, I see my mistake now, thanks. I see from the answers that there is a simpler way to calculate this. –  Jul 27 '16 at 07:04

2 Answers2

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Hint: $$\dfrac{d}{dt} \sum_{n=0}^\infty t^n = \sum_{n=1}^\infty n t^{n-1}$$ for $|t|<1$.

Robert Israel
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    To use that I need to show that there is uniform convergence on $[0,1] $ ? because this sum only converge uniformly on $[a,\infty) $ for $a\gt0 $ –  Jul 27 '16 at 07:07
  • The series in my answer above converge uniformly on $[-1+\epsilon, 1-\epsilon]$ for any $\epsilon > 0$. Not on $[0,1]$, but you don't need it at $1$. – Robert Israel Jul 27 '16 at 15:11
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The RHS of your first inequlity should probably be $ 1\over \exp 1$ rather than $n\over \exp n$. By the way, let $q= \exp -x$, the general term of your series is $xnq^n$, so it converges either if $x=0$ or if $0\leq q <1$, i.e. $x>0$. You probably know that $\sum _{k=1} ^\infty kq^k= {q\over (1-q)^2}$, and $\sum _{k=1}^\infty {kx \over \exp kx}={x \exp -x \over (1-\exp -x)^2}$

Thomas
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