# Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$

How can we find the exact value of the infinite sum $$\displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}?$$

This problem appears in:

-
If I have interpreted your question correctly, the series diverges. – Sabyasachi Mar 7 '14 at 16:44
@Tyler: All members of the series are less than $1$. – triple_sec Mar 7 '14 at 16:45
As it is written, the taking of the fractional part is redundant: $0 < e - (1+1/n)^n < 1$ for all $n \ge 1$. – heropup Mar 7 '14 at 17:01
@k1.M I guess because $(1+1/n)^n$ fails to converge to $e$ “quickly enough.” – triple_sec Mar 7 '14 at 17:03
On a related note, $\quad\displaystyle\sum_{n=1}^\infty\bigg(e-\sum_{k=0}^n\frac1{k!}\bigg)=1.$ – Lucian Mar 7 '14 at 17:15

We have the following approximation for $\big(1+\frac{1}{n}\big)^n$: $$\left(1+\frac{1}{n}\right)^{\!n}=\mathrm{e}^{n\log(1+\frac{1}{n})}=\mathrm{e}^{1-\frac{1}{2n}+{\mathcal O}(n^{-2})}=\mathrm{e}\left(1-\frac{1}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\Big)\right),$$ since $$\log \Big(1+\frac{1}{n}\Big)=\frac{1}{n}-\frac{1}{2n^2}+{\mathcal O}\Big(\frac{1}{n^3}\Big) \quad\text{and}\quad \mathrm{e}^h=1+h+{\mathcal O}(h^2),$$ for $h$ small and $n$ large. Hence $$\mathrm{e}-\left(1+\frac{1}{n}\right)^{\!n}=\frac{\mathrm{e}}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\!\Big).$$ This implies that the series diverges, i.e., $$\sum_{n=1}^\infty \left\{\mathrm{e}-\left(1+\frac{1}{n}\right)^{\!n}\right\}=\infty.$$ Note. All the term of the sequence $$a_n=\mathrm{e}-\Big(1+\frac{1}{n}\!\Big)^n, \quad n\in\mathbb N,$$ are positive, since $\mathrm{e}^{1/n}>1+\frac{1}{n}$. It is noteworthy that the series $\sum_{n=1}^\infty \big\{ \mathrm{e}-\big(1+\frac{1}{n}+\frac{1}{2n^2}\!\big)^n\big\}$, also of positive terms is convergent.
@Yiorgos S. Smyrlis,Note that you must also prove that every element of the series is positive,i.e.,$$0<e-(1+\frac 1n)^n$$ – k1.M Mar 10 '14 at 9:04
@k1.M, it follows immediately from the equality $e - (1+1/n)^n = e/(2n) + O(1/n^2)$ that $e - (1+1/n)^n$ is positive for $n$ large enough. – Antonio Vargas Mar 13 '14 at 14:50
Alternatively, without using logs, note that if we expand $( 1+ \frac{1}{n})^{n}$ via the binomial theorem, for $2 \leq k \leq n,$ the $k$-th summand is $\frac{1}{k!}(1 - \frac{1}{n}) \ldots (1 - \frac{(k-1)}{n}) \leq \frac{1}{k!}(1 - \frac{1}{n}) .$ This means that $e - (1+\frac{1}{n})^{n}$ is greater than or equal to $\frac{1}{n}\sum_{k=2}^{n} \frac{1}{k!}.$ This is at least $\frac{1}{2n}$ for $n \geq 2.$ Hence the given sum is greater than or equal to $\sum_{n=2}^{\infty} \frac{1}{2n},$ which diverges.