0
$\begingroup$

I know that the Cauchy product is defined $$\left(\sum_{n=1}^\infty\frac{\log n}{e^n}\right)\left( \sum_{n=1}^\infty\frac{1}{e^n} \right)= \sum_{n=1}^\infty\sum_{k=1}^n\frac{\log k}{e^{k+n-k+1}},$$ and from this compute $$\sum_{n=1}^\infty \frac{\log n}{e^n}= \left( 1-\frac{1}{e} \right)\sum_{n=1}^\infty\frac{\log n!}{e^n}. $$

My purpose was to state a nice identity, but I have problems. My computations were by partial summation $$\sum_{n\leq x}\frac{\log n}{e^n}=[x]\frac{\log x}{e^x}-\int_1^x [t]\frac{\frac{1}{t}-\log t}{e^t}dt,$$ where the first summand in RHS vanishes as $x\to\infty$, and the integral converges. If there are no mistakes, I know write this last as the series of $$k\int_k^{k+1}\frac{\frac{1}{t}-\log t}{e^t}dt.$$

But I don't know if it is possible continue, because Wolfram Alpha say that $\sum_{n\leq x}\frac{\log n}{e^n}$ is equal to $$-PolyLog^{(1,0)}(0,\frac{1}{e}).$$

I don't know if such exercise is in the literature, I ask it

Question. Can you compute $$\sum_{n=1}^\infty\frac{\log n!}{e^n}$$ or explain what is $$-PolyLog^{(1,0)}(0,\frac{1}{e})$$ (I say a understandable explanation, how is defined this last special function, how to evaluate this particular value and why converges it) to get an identity? Thanks in advance.

$\endgroup$
0
$\begingroup$

The polylogarithm function is defined as $$\textrm{Li}_{s}\left(z\right)=\sum_{k\geq1}\frac{z^{k}}{k^{s}} $$ for all complex $s$ and for $\left|z\right|<1 $. So observe that $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}k^{s}} $$ and so $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)_{s=0}=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}} $$ and this is the $PolyLog^{\left(1,0\right)}\left(0,\frac{1}{e}\right)$ of Wolfram alpha. So the parenthesis $(1,0)$ indicates the differentiation with respect to $s$ one time and no differentiation with respect to $z$.

$\endgroup$
  • $\begingroup$ Now I understad this special function, I believe that the purpose of my question is satisfied, because I unerstand more better previous Cauchy product. Very thanks much for the lesson. $\endgroup$ – user243301 May 5 '16 at 21:00

Your Answer

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