1
$\begingroup$

Does there exist a closed-form expression for the series below?

$$\sum_{r=1}^{\infty} \frac{e^{-rx}}{\ln(r+1)}$$

I’m pretty sure this series converges for $x \geq 0$, though I haven’t checked myself. Even so, I’m not sure that there’s even an expression for this. Just wondering if there was.

$\endgroup$
2
$\begingroup$

Consider $\;z:=e^{-x}\,$ for $\,x>0\,$ (i.e. $\;z<1$) then you want : $$S_z:=\sum_{k=1}^{\infty} \frac{z^k}{\ln(k+1)}$$ Since $\;\displaystyle \frac 1{\ln(k+1)}=\int_0^\infty (k+1)^{-s} ds\ \;$ for $k>0\;$ we may rewrite $S_z$ as : $$S_z:=\sum_{k=1}^{\infty} \int_0^\infty \frac{z^k}{(k+1)^s}\,ds$$ I'll let you prove that $\sum$ and $\int$ may be exchanged to get : \begin{align} S_z&=\frac 1z\int_0^\infty \sum_{k=1}^{\infty} \frac{z^{k+1}}{(k+1)^s}\,ds\\ &=\frac 1{z}\int_0^\infty\operatorname{Li_s}(z)-z\;ds\\ \end{align} with $\operatorname{Li_s}$ the polylogarithm function (evaluation using alpha).

Of course this doesn't help much for a closed form. Integrating polylogarithms relatively to their first parameter (denominator's power) is not easier than integrating zeta functions but it could define a new (possibly interesting) special function.
Analytic continuation may be used too to evaluate your series for any value of $z\neq 1$.

$\endgroup$
  • $\begingroup$ Holy I didn’t see that coming. Nice super awesome thanks for this beautiful answer. $\endgroup$ – Horus Mar 13 '18 at 12:35
  • $\begingroup$ On a side note can’t this already be considered an analytic continuation already? Seeing that I could sub x as any value not equal to zero, doesn’t this already extend it to the rest of the complex values? $\endgroup$ – Horus Mar 13 '18 at 13:03
  • 1
    $\begingroup$ Glad you liked this answer @Horus and yes for $|z|>1$ we may use the $\;\operatorname{Li}_s(z)/z-1\;$ integral for the analytic continuation even if the definition of $\,S_z\,$ and the series defining the polylogarithm $\;\displaystyle \sum_{k=1}^{\infty} \frac{z^k}{k^s}\;$ are divergent. Excellent continuation, $\endgroup$ – Raymond Manzoni Mar 13 '18 at 13:56
  • 1
    $\begingroup$ That’s another upvote for me haha $\endgroup$ – Horus Mar 13 '18 at 15:03
1
$\begingroup$

It's rather unlikely to have a closed form. For $x > 0$, it does converge: just use the comparison test with a geometric series. For $x=0$, it does not converge.

$\endgroup$
  • $\begingroup$ Wouldn’t it also converge for $x=0$ too since using the integral test gives it a finite value? $\endgroup$ – Horus Mar 5 '18 at 20:05
  • $\begingroup$ No. $\int_1^\infty \frac{dr}{\ln(r+1)}$ diverges. $\endgroup$ – Robert Israel Mar 5 '18 at 20:07
  • $\begingroup$ Wait no sorry I remembered the integral wrong. Yes it doesn’t converge at zero. $\endgroup$ – Horus Mar 5 '18 at 20:07
  • $\begingroup$ It does converge at 0, but not at infinity. $\endgroup$ – user Mar 5 '18 at 20:09
  • $\begingroup$ Seriously? Based on integral test for convergence it should diverge at 0 but converge to zero at infinity no? @user $\endgroup$ – Horus Mar 5 '18 at 20:16

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.