The target is to find the upper bound of the summation.

I can only get that: since $\sqrt{i}<i$,
$$S < \sum \limits_{i=1}^n e^{-i}i = \frac{1-e^{-n}}{e(1-e^{-1})^2} - \frac{n}{e^{(n+1)}(1-e^{-1})} = T$$

then, we have: $$ \lim \limits_{n\to\infty} T = \frac{1}{e(1-e^{-1})^2} \approx 0.9206735942077925 $$.

I use python to get that S converges to 0.707240718486804.

Does anyone has a more tight upper bound than $\frac{1}{e(1-e^{-1})^2} $? Thanks a lot.

  • 2
    $\begingroup$ Well, it's obvious you can get, for example, $T - (\frac{2 - \sqrt{2}}{e})$. Also may be help that the sum of series is polylogarithm $\operatorname{Li}_\frac{1}{2}(\frac{1}{e})$. $\endgroup$
    – mihaild
    Commented Apr 8, 2019 at 16:20
  • $\begingroup$ Thanks a lot. I have just heard about Li function but never have learned. I was wondering that: can $Li_{\frac 12}(\frac 1e)$ be represented by some fundamental math symbols? (I guess it cannot because maybe Li function is just like $\pi$ and $e$?) $\endgroup$ Commented Apr 9, 2019 at 15:22

1 Answer 1


Besides expressing in polylogarithm form, a slightly tighter bound is \begin{align*} \left(\sum_{i=1}^{\infty} e^{-i}\sqrt{i}\right)^2 &= \left(\sum_{i=1}^{\infty} e^{-i/2}e^{-i/2}\sqrt{i}\right)^2 \\ &\le \left(\sum_{i=1}^{\infty} e^{-i}\right)\left(\sum_{i=1}^{\infty} e^{-i}i\right) && \text{Cauchy-Schwarz}\\ &= \frac{1}{e-1}T \\ &\approx 0.5358 \\ \implies \sum_{i=1}^{\infty} e^{-i}\sqrt{i} &\le 0.7320 \end{align*} There are an infinite number of ways to make this bound tighter by simply writing out the first few terms and applying Cauchy-Schwarz on the rest: \begin{align*} \sum_{i=1}^{\infty} e^{-i}\sqrt{i} &= e^{-1} + \sum_{i=2}^{\infty} e^{-i}\sqrt{i} \\ &\le e^{-1} + \left(\sum_{i=2}^{\infty} e^{-i}i\right)^{1/2}\left(\sum_{i=2}^{\infty} e^{-i}\right)^{1/2} \\ &= e^{-1} + \frac{\sqrt{2e-1}}{(e-1)^{3/2}e} \\ &\approx 0.7119 \end{align*}

  • $\begingroup$ Thanks a lot! I am wondering that: is it possible to get the exact upper bound by some fundamental math symbols? (I saw $Li_{\frac 12}(\frac 1e)$ representations before.) $\endgroup$ Commented Apr 9, 2019 at 15:26

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