I have been having some trouble trying to find a closed form for this sum. It seems to converge really slowly.

Find a closed form for $$S=\sum_{n=1}^\infty\left[e-\left(1+\dfrac{1}{n}\right)^n\right].$$

All I got so far is

$$ \begin{align} e-\left(1+\dfrac{1}{n}\right)^{n} & = \sum_{k=0}^\infty\frac{1}{k!} -\sum_{k=0}^n\binom{n}{k}\frac{1}{n^{k}} \\ & = \sum_{k=0}^\infty\frac{1}{k!}\left(1-\dfrac{n!}{(n-k)!}\dfrac{1}{n^k}\right) \\ & = \sum_{k=0}^\infty\frac{1}{k!}\left(1-\dfrac{(n)_k}{n^k}\right) \\ \end{align}, $$

$$ S=\sum_{k=0}^\infty\frac{1}{k!}\sum_{n=1}^\infty\left(1-\dfrac{(n)_k}{n^k}\right). $$

Where $(n)_k$ is the Pochhammer symbol. But I don not know how I could carry on from here.

  • 3
    $\begingroup$ Is there a reason you think there is a closed form? $\endgroup$ – Thomas Andrews Mar 21 '15 at 22:31
  • $\begingroup$ Also, is there a reason you didn't use MathJax, but posted an image? $\endgroup$ – Thomas Andrews Mar 21 '15 at 22:37
  • $\begingroup$ Well, actually there is no particular reason why I´d think there is a closed form for this, but I felt curious about it.... and sorry, but it´s difficult form me using MathJax... I hope that´s not a problem $\endgroup$ – Joseph Mar 21 '15 at 22:47
  • $\begingroup$ In the title, you have an "over $n$"; in the picture, you don't. Please edit so things match. $\endgroup$ – Gerry Myerson Mar 21 '15 at 23:11
  • $\begingroup$ @GerryMyerson That "over $n$" I assumed meant "summing over $n$." As in $\sum_{n=1}^\infty\dots$. Certainly, not written that way in his answer. $\endgroup$ – Thomas Andrews Mar 21 '15 at 23:13

$+\infty$ is a nice closed form.

By the Hermite-Hadamard inequality we have: $$\log\left(1+\frac{1}{n}\right)^n = n\int_{n}^{n+1}\frac{dx}{x}\leq\frac{n}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)= 1-\frac{1}{2n+2}$$ hence, by the concavity of $1-e^{-x}$ over $\left[0,\frac{1}{4}\right]$: $$ e-\left(1+\frac{1}{n}\right)^n \geq e\left(1-e^{-1/(2n+2)}\right)\geq\frac{4e}{2n+2}(1-e^{-1/4})\geq\frac{6}{5}\cdot\frac{1}{n+1}. $$ We can also prove that for any $n\geq 1$

$$ e-\left(1+\frac{1}{n}\right)^n \geq \frac{e}{2n+2}$$

holds. The conclusion is just the same.

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  • $\begingroup$ I have a proof, similar to my proof here, that the difference is $\le\frac e{2n+1}$, but the only proof I've found for the highlighted inequality is not very simple. Do you have a simple proof of that inequality? $\endgroup$ – robjohn Mar 23 '15 at 21:49

Here's yet another approach, using the inequality $e^x\ge1+x$.

$e^x\ge1+x$ implies that $\log(1+x)\le x$. Thus, $$ \begin{align} \log\left(1+\frac1k\right)-\log\left(1+\frac1{k+1}\right) &=\log\left(1+\frac1{k(k+2)}\right)\\ &\le\frac1{k(k+2)}\\ &=\frac12\left(\frac1k-\frac1{k+2}\right) \end{align} $$ Therefore, $$ \begin{align} n\log\left(1+\frac1n\right) &=n\sum_{k=n}^\infty\left[\log\left(1+\frac1k\right)-\log\left(1+\frac1{k+1}\right)\right]\\ &\le\frac n2\sum_{k=n}^\infty\left(\frac1k-\frac1{k+2}\right)\\ &=\frac n2\left(\frac1n+\frac1{n+1}\right)\\ &=1-\frac1{2n+2} \end{align} $$ Exponentiating and applying $e^x\ge1+x$ yields $$ \begin{align} \left(1+\frac1n\right)^n &\le e\cdot e^{-\frac1{2n+2}}\\ &\le\frac e{1+\frac1{2n+2}}\\ &=e\left(1-\frac1{2n+3}\right) \end{align} $$ Therefore, $$ \bbox[5px,border:2px solid #00A0F0]{e-\left(1+\frac1n\right)^n\ge\frac e{2n+3}} $$ Of course, this leads to the same conclusion: divergence of the series.

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A more elementary answer to Jack's nice but perhaps complex answer involves looking at just the case $k=2$.

Writing $(n)_k=n(n-1)\cdots(n-(k-1))$, the falling factorial, we have:

$$\begin{align} e-(1+1/n)^n &= \sum_{k=0}^\infty \frac{1}{k!}\left(1-\frac{(n)_k}{n^k}\right)\\ &\geq \frac{1}{2n} \end{align}$$

since all the terms in the sum are positive, and $\frac{1}{2n}$ is the term when $k=2$.

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  • $\begingroup$ Thank you guys! I wrote a Matlab algorithm to evaluate the sum and it seems to grow to infinity, but very slowly. $\endgroup$ – Joseph Mar 21 '15 at 23:37
  • $\begingroup$ Yes, probably about like some multiple of $\log n$. $\endgroup$ – Thomas Andrews Mar 21 '15 at 23:38
  • $\begingroup$ @ThomasAndrews Nice! +1. $\endgroup$ – Olivier Oloa Mar 22 '15 at 19:37


The first terms of the Taylor expansion of $e-(1+x)^{1/x}$ are


As there are no singularities around the origin, the series converges for $|x|\le\frac12$.

Summing with $x=\dfrac1n$, your series is asymptotic to the harmonic one.

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  • $\begingroup$ I am leaving this as a hint because it is lacking rigor. $\endgroup$ – Yves Daoust Mar 23 '18 at 10:21

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