TITLE: Series$\sum_{n=1}^\infty\big((1+1/n)^n-e\big)$
QUESTION [9 upvotes]: I can't figure out whether this series is convergent.
I'm trying to use d'Alembert or Cauchy ratio tests, but however far I go with Taylor series it always ends up being one.
The series is :
$$\sum_{n=1}^\infty \big((1+1/n)^n-e\big)$$

REPLY [2 votes]: As an answer hasn't been accepted yet, I assume that there's something left to be desired. So I'll take a different stab at the problem that only involves calculus.
We first observe, as in the comments, that
$$\Bigl(1+\frac{1}{n}\Bigr)^n-e=e^{n\ln(1+1/n)}-e\,.$$
Elementary calculus arguments can show that the function $x\ln(1+1/x)$ is monotone increasing on $(0,\infty)$ and that its range has $1$ as a supremum and has $0$ as an infimum.
We now invoke the Mean Value Theorem to conclude that for every $n\in\mathbb{N}$ there is a real $\xi_n$ between $1$ and $n\ln(1+1/n)$ such that
$$e-\Bigl(1+\frac{1}{n}\Bigr)^n=e^{\xi_n}\bigl(1-n\ln(1+1/n)\bigr)\,.$$
From what we know about $x\ln(1+1/x)$, we can conclude that
$$e^0\bigl(1-n\ln(1+1/n)\bigr)\leq e-\Bigl(1+\frac{1}{n}\Bigr)^n\leq e^1\bigl(1-n\ln(1+1/n)\bigr)\,.$$
Now we use the inequalities
\begin{equation}\frac{1}{2(1+x)}\leq 1-x\ln(1+1/x)\leq\frac{1}{x+1}\end{equation}
which is true for all positive $x$ to conclude
$$\frac{1}{2(1+n)}\leq e-\Bigl(1+\frac{1}{n}\Bigr)^n\leq\frac{e}{1+n}$$
which will carry the rest of the argument.
The real crux to this argument are the inequalities
\begin{equation}\frac{1}{2(1+x)}\leq 1-x\ln(1+1/x)\leq\frac{1}{x+1}\,.\end{equation}
The right-hand inequality is equivalent to the more familiar inequality
$$\ln(1+x)\leq x$$
which I can derive if needed.
The left-hand inequality is equivalent to the slightly stronger inequality
$$\ln(1+x)\leq x-\frac{x^2}{2(1+x)}$$
which is however established in very much the same way as the weaker inequality.