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

Solution 1:

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.

Solution 2:

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.