How find the limit of $$\lim_{n\rightarrow\infty}\sum_{i=1}^n \ln(1+\frac{1}{e^i})$$

I just prove this limit has upper boundary $\frac{e-1}{e}$ and lower boundary $\ln(\frac{e-1}{e})$.

Can any one solve it?

  • 1
    $\begingroup$ Are you sure there is a "nice" expression for it (cf. Wolfram? Where did that question arise? $\endgroup$ – Clement C. May 10 '16 at 14:22

$$\sum_{n\geq 1}\log(1+e^{-n})=\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m}e^{-mn} =\sum_{m\geq 1}\frac{(-1)^{m+1}}{m(e^m-1)}$$ where the last series is a fast-convergent series with alternating signs, whose terms are decreasing in absolute value. It follows that the LHS is less than $\frac{1}{e-1}$ but greater than $\frac{1}{e-1}\left(1-\frac{1}{2e+2}\right)$. I would not bet on a nice closed form.

  • $\begingroup$ Maybe not my problem, but if I may -- why was this downvoted? $\endgroup$ – Clement C. May 10 '16 at 14:26

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