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:
T. Andreescu, T. Radulescu & V. Radulescu, Problems in Real Analysis: Advanced Calculus on the real line, p.114.
 A: 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.
A: 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.
