# A closed form for $\sum\limits_n(e-(1+1/n)^n)$

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.

• Is there a reason you think there is a closed form? – Thomas Andrews Mar 21 '15 at 22:31
• Also, is there a reason you didn't use MathJax, but posted an image? – Thomas Andrews Mar 21 '15 at 22:37
• 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 – Joseph Mar 21 '15 at 22:47
• In the title, you have an "over $n$"; in the picture, you don't. Please edit so things match. – Gerry Myerson Mar 21 '15 at 23:11
• @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. – 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.

• 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? – 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.

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 $k=2$.

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

Hint:

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

$$\frac{ex}2-\frac{e11x^2}{24}+\frac{e7x^3}{16}-\frac{e2447x^4}{5760}+\cdots$$

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.

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