I want to evaluate $$\sum_{n=1}^{\infty} \frac{e - \left(1 + \frac{1}{n}\right)^n}{\sqrt{n}}$$ But I'm not sure how to approach it. Mathematica suggests that it converges pretty slowly and gives something like 2.57... after around 20,000 terms, but then starts to choke.

To see that it converges, I think I can write the following: $$\left(1 + \frac{1}{n}\right)^n = e^{n \ln\left(1 + \frac{1}{n}\right)} = e^{n\left(\frac{1}{n} - \frac{1}{2n^2} + O(n^{-3})\right)} = ee^{-\frac{1}{2n}}e^{O(n^{-2})} $$ and note that $e^{O(n^{-2})} \to 1$ as $n \to \infty$, so that $$ e - \left(1 + \frac{1}{n}\right)^n \approx e - ee^{-\frac{1}{2n}} = e(1 - e^{-\frac{1}{2n}}) = e\left(\frac{1}{2n} + O(n^{-2})\right) = O(n^{-1})$$

and we have $$\frac{e - \left(1 + \frac{1}{n}\right)^n}{\sqrt{n}} = O(n^{-3/2})$$ which says this is asymptotically just a p-series.

Any ideas?

  • 5
    $\begingroup$ There might be no known closed form $\endgroup$
    – Amr
    Commented Feb 16, 2013 at 18:29
  • 1
    $\begingroup$ The step using that $e^{O(1/n^2)}\to1$ to deduce that $e-(1+1/n)^n\sim e-ee^{-1/(2n)}$ is wrong (the conclusion holds but the argument fails). $\endgroup$
    – Did
    Commented Oct 11, 2014 at 7:31

3 Answers 3


First, excellent job on estimating the summand - few people are skilled with power-series manipulations like that.

If you expand the summand into a power series (at $n=\infty$), you'll get something of the form $$ \frac{e-(1+\frac1n)^n}{\sqrt n} = \frac e2n^{-3/2} - \frac{11e}{24}n^{-5/2} + \frac{7e}{16}n^{-7/2} + \cdots = \sum_{k=0}^\infty c_k n^{-k-3/2} $$ for certain easily computed constants $c_k$. Choosing whatever truncation points $K$ and $N$ you like, you can then write \begin{align*} \sum_{n=1}^\infty \frac{e-(1+\frac1n)^n}{\sqrt n} &= \sum_{n=1}^\infty \bigg( \sum_{k=0}^K c_k n^{-k-3/2} + \bigg( \frac{e-(1+\frac1n)^n}{\sqrt n} - \sum_{k=0}^K c_k n^{-k-3/2} \bigg) \bigg) \\ &= \sum_{k=0}^K c_k \zeta(k+3/2) + \sum_{n=1}^N \bigg( \frac{e-(1+\frac1n)^n}{\sqrt n} - \sum_{k=0}^K c_k n^{-k-3/2} \bigg) + \sum_{n>N} O(n^{-K-5/2}). \end{align*} Everything other than the error term can be calculated to as many decimal places as you want (and the implicit constant in the $O$-notation could also be calculated, if you want to be completely rigorous). With this accelerated convergence, I find the value of your original sum to be $2.5912775703968337633$, probably accurate to that many decimal places.

(Then, if you want, you can try an inverse constant lookup to see if that value matches any closed form - although I doubt it will in this case.)

  • $\begingroup$ I've added a bounty in the (probably vain) hope that there's a closed form, but if no one finds one, I'll just give it to you instead. $\endgroup$
    – AndrewG
    Commented Feb 18, 2013 at 22:06
  • $\begingroup$ "few people are skilled with power-series manipulations like that." excuse me, but there are a lot of 'smart' series manipulators on this site. I wouldn't say the expansion given is of excellence. :-) $\endgroup$ Commented Aug 14, 2017 at 0:19

Here we find another representation of the sum with improved convergence properties.

We have $$\begin{eqnarray*} e - \left(1 + \frac{1}{n}\right)^n &=& \sum_{k=0}^\infty \frac{1}{k!} - \sum_{k=0}^n {n\choose k} \frac{1}{n^k} \\ &=& \sum_{k=0}^\infty \frac{1}{k!} \left(1-\frac{n!}{(n-k)!}\frac{1}{n^k}\right) \\ &=& \sum_{k=1}^\infty \frac{1}{k!} \left(1-\frac{n^{\underline k}}{n^k}\right), \end{eqnarray*}$$ where $n^{\underline k}$ is the falling factorial. But $$n^{\underline k} = \sum_{j=1}^k (-1)^{k-j} \left[ k\atop j\right] n^j,$$ where $\left[ k\atop j\right]$ is the unsigned Stirling number of the first kind. Thus, $$\begin{eqnarray*} e - \left(1 + \frac{1}{n}\right)^n &=& -\sum_{k=2}^\infty \frac{1}{k!} \sum_{j=1}^{k-1} (-1)^{k-j} \left[ k\atop j\right] \frac{1}{n^{k-j}}, \end{eqnarray*}$$ and so $$\begin{eqnarray*} \sum_{n=1}^{\infty} \frac{e - \left(1 + \frac{1}{n}\right)^n}{\sqrt{n}} &=& \sum_{k=2}^\infty \underbrace{\frac{1}{k!} \sum_{j=1}^{k-1} (-1)^{k-j+1} \left[ k\atop j\right] \zeta(k-j+\textstyle \frac{1}{2})}_{a_k}. \end{eqnarray*}$$

Below we give the partial sums to 50 digits. $$\begin{array}{cl} N & \sum_{k=2}^N a_k \\ \hline 10 & 2.5912768743966448667041943446771083714079106923575 \\ 20 & 2.5912775703968337622590116959919424146275651059115 \\ 30 & 2.5912775703968337632934326247597530975932248904119 \\ 40 & 2.5912775703968337632934326247597625731591505820979 \\ 50 & 2.5912775703968337632934326247597625731591505821009 \end{array}$$

  • $\begingroup$ I don't think that's 50 digits, but rather, 50 terms. $\endgroup$ Commented Aug 14, 2017 at 0:21
  • $\begingroup$ @SimplyBeautifulArt: The last result has 50 decimal digits and 49 (or, depending on how you count, 1225) terms. $\endgroup$
    – user26872
    Commented Aug 14, 2017 at 16:33

Start with the power series for $n\log\left(1+\frac1n\right)\color{#C00000}{-1}$: $$ -\frac1{2n}+\frac1{3n^2}-\frac1{4n^3}+\frac1{5n^4}-\frac1{6n^5}+\dots\tag{1} $$ Apply the power series for $\color{#C00000}{1}-e^x$ to $(1)$ and divide by $\sqrt{n}$: $$ \frac1{2n^{3/2}}-\frac{11}{24n^{5/2}}+\frac7{16n^{7/2}}-\frac{2447}{5760n^{9/2}}+\dots\tag{2} $$ Finally, apply the Euler-Maclaurin Sum Formula to $(2)$: $$ \frac Se-\frac1{n^{1/2}}+\frac5{9n^{3/2}}-\frac7{15n^{5/2}}+\frac{4391}{10080n^{7/2}}-\dots\tag{3} $$ Due to the $\color{#C00000}{-1}$ in $(1)$ and the $\color{#C00000}{1}$ in $(2)$ instead of $e$, $(3)$ needs to be multiplied by $e$ to approximate $$ \sum_{k=1}^n\frac{e-\left(1+\frac1k\right)^k}{\sqrt{k}}\tag{4} $$ Using $(3)$ out to the $n^{-35/2}$ term and $(4)$ with $n=10000$, we get $$ S=\small 2.59127757039683376329343262475976257315915 05821009\ 381932479091255849847307 $$ which agrees with user26872's answer.

Asymptotic Expansion

Here is the expansion started in $(3)$: $$ \begin{align} &\frac1e\sum_{k=1}^n\frac{e-\left(1+\frac1k\right)^k}{\sqrt{k}}\\[4pt] &=\frac Se-\frac1{n^{1/2}}+\frac5{9n^{3/2}}-\frac7{15n^{5/2}}+\frac{4391}{10080n^{7/2}}-\frac{21949}{51840n^{9/2}}+\frac{6656599}{15966720n^{11/2}}\\[4pt] &-\frac{397567}{967680n^{13/2}}+\frac{528603833}{1306368000n^{15/2}}-\frac{271019261}{676823040n^{17/2}}+\frac{1395775243843}{3494795673600n^{19/2}}\\[4pt] &-\frac{3075041160157}{7725337804800n^{21/2}}+\frac{13560289331497903}{34648140054528000n^{23/2}}-\frac{25805885669173201}{66952927641600000n^{25/2}}\\[4pt] &+\frac{3117317490074490877}{7809389480116224000n^{27/2}}-\frac{6994535029664135647}{16775725549879296000n^{29/2}}\\[4pt] &+\frac{1574300404641490648151}{4572831395579166720000n^{31/2}}-\frac{6450883875238096571903}{28322052514554839040000n^{33/2}}\\[4pt] &+\frac{3169495620834596471965948267}{4746209560389099926323200000n^{35/2}} \end{align} $$


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