Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
There might be no known closed form –  Amr Feb 16 '13 at 18:29

2 Answers 2

up vote 7 down vote accepted
+50

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.)

share|improve this answer
    
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. –  AndrewG Feb 18 '13 at 22:06

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. $$\begin{array}{cl} N & \sum_{k=2}^N a_k \\ \hline 10 & 2.5912768743966448667041943446771083714079106923575 \\ 20 & 2.5912775703968337622590116959919424146275651059115 \\ 30 & 2.5912775703968337632934326247597530975932248904119 \\ 40 & 2.5912775703968337632934326247597625731591505820979 \\ 50 & 2.5912775703968337632934326247597625731591505821009 \end{array}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.