Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a  way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by induction that $\big(1+\frac{1}{n}\big)^n\le3-\frac{1}{n}$, as in
$\hspace{.4 in}$ A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?

As suggested by Clement C., let:
$$ f(x)=\left(1+\frac{1}{x}\right)^{x}+\frac{1}{x}.\tag{1}$$
Then:
$$ f'(x) = \left(1+\frac{1}{x}\right)^{x}\left(\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right)-\frac{1}{x^2}\tag{2} $$
but, due to convexity:
$$\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}=-\frac{1}{x+1}+\int_{x}^{x+1}\frac{dt}{t}\geq \frac{1}{2(x+1)^2}\tag{3}$$
hence for any $x\geq 8$:
$$ f'(x)\geq \frac{\left(1+\frac{1}{8}\right)^8}{2(x+1)^2}-\frac{1}{x^2}>0.\tag{4} $$
