Series$\sum_{n=1}^\infty\big((1+1/n)^n-e\big)$ 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 :
$$\sum_{n=1}^\infty \big((1+1/n)^n-e\big)$$
 A: Take the negative of that series to make the terms positive and limit compare it with $\frac{1}{n}$
$$\lim_{n\to\infty} \frac{e-\left(1+\frac{1}{n}\right)^n}{\frac{1}{n}} \to \frac{0}{0}$$
$$\implies \lim_{n\to\infty} \frac{-\left(1+\frac{1}{n}\right)^n\Biggr(\log\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\Biggr)}{-\frac{1}{n^2}}$$
which we get from L'Hopital and logarithmic differentiation. If this limit exists, we can decompose it into a product. The limit on the left is $e$, so let's evaluate
$$\lim_{n\to\infty} \frac{\log\left(1+\frac{1}{n}\right)-\frac{1}{n+1}}{\frac{1}{n^2}}\to \frac{0}{0}$$
$$\implies \lim_{n\to\infty} \frac{\frac{1}{1+\frac{1}{n}}\cdot\left(-\frac{1}{n^2}\right)+\frac{1}{(n+1)^2}}{-\frac{2}{n^3}} = \lim_{n\to\infty}\frac{1}{2}\frac{n^2}{(n+1)^2} = \frac{1}{2}$$
This means the original limit equals $\frac{e}{2}$, so the series behaves as $\frac{1}{n}$, i.e. it diverges by the limit comparison test.
A: 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.
A: Let $u=1/n$ then the Mc Laurin expansion is $$(1+u)^{1/u}-e-eu/2+11eu^2/24+...$$
Then $$\sum_{n=1}^{\infty} [\left(1+\frac{1}{n}\right)^n-e]=\sum_{n=1}^{\infty} \left( e-\frac{e}{2n}+\frac{11 }{24 n^2}-...-\frac{e}{2} \right)=-\frac{e}{2}\sum_{n=1}^{\infty} \frac{1}{n}+ \frac{11e}{24}\sum_{n=1}^{\infty}\frac{1}{n^2}+... $$ $$\sim -\frac{e}{2} \sum_{1}^
{\infty} \frac{1}{n} = -\infty$$ Here the very first series being divergent the given sum diverges to $-\infty$.
We may also write that $$\sum_{n=1}^{N}\left[\left (1+\frac{1}{n} \right)^n-e \right] \sim \frac{-e}{2} H_N \sim \frac{-e}{2} \ln N.$$ 
where $H_n$ is the sum of the Harmonic series.  Further, you may see the plot of $$S(N)=\sum_{n=1}^{N}\left[\left (1+\frac{1}{n} \right)^n-e \right]$$ vs $N$ below
$S(N)$">
