TITLE: Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$
QUESTION [28 upvotes]: 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.

REPLY [8 votes]: 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.