Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing

Let $$ \eqalign{f(n) = \dfrac{1}{n} + \left( 1 + \dfrac{1}{n}\right)^n &= \dfrac{1}{n} + \exp\left( n \ln\left(1+\dfrac{1}{n}\right)\right) \cr &= \dfrac{1}{n} + \exp\left(1 - \dfrac{1}{2n} + \dfrac{1}{3n^2} + O\left(\dfrac{1}{n^3}\right)\right) \cr &= e - \dfrac{e-2}{2n} + \dfrac{11e}{24 n^2} + O\left(\dfrac{1}{n^3}\right) }$$

Then $$\eqalign{f(n+1) &= e - \dfrac{e-2}{2n+2} + \dfrac{11e}{24 (n+1)^2} + O\left(\dfrac{1}{n^3}\right)\cr &= e - \dfrac{e-2}{2n} + \dfrac{23 e - 24}{24 n^2} + O\left(\dfrac{1}{n^3}\right) \cr f(n+1) - f(n) &= \dfrac{12e-24}{24n^2} + O\left(\dfrac{1}{n^3}\right)}$$

and since $e > 2$, this is positive for sufficiently large $n$.

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 $x\geq 8$: $$ f'(x)\geq \frac{\left(1+\frac{1}{8}\right)^8}{2(x+1)^2}-\frac{1}{x^2}>0.\tag{4} $$

Let $a_n = (1 + 1/n)^n.$

We want to show $a_{n+1} - a_{n} \geq \dfrac{1}{n(n+1)}$ for large $n$.

$\dfrac{a_{n+1}}{a_n} = \left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1}.$

The RHS can be expanded as

$\left(1 + \dfrac{1}{n}\right) \left(1 - \dfrac{1}{(n+1)^2}\right)^{n+1} = \dfrac{n+1}{n} \times \left( \underbrace{1 - \dfrac{1}{n+1}} + \dfrac{1}{2!(n+1)^2}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} - \dfrac{1}{3!(n+1)^3}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) + \dots (-1)^{n+1} \dfrac{1}{(n+1)!(n+1)^{n+1}}\underbrace{\left(1 - \dfrac{1}{n+1}\right)} \left(1 - \dfrac{2}{n+1} \right) \dots\left(1 - \dfrac{n}{n+1}\right)\right).$

Since $ \dfrac{n+1}{n} \times (1-\dfrac{1}{n+1}) = 1$, we have $\dfrac{a_{n+1}}{a_n} = 1 + \dfrac{1}{2!(n+1)^2} - \dfrac{1}{3!(n+1)^3} \left(1 - \dfrac{2}{n+1} \right) + \dots$

So,

$|\dfrac{a_{n+1}}{a_n} - 1 - \dfrac{1}{2(n+1)^2}| \leq \dfrac{1}{3!(n+1)^3} + \dfrac{1}{4!(n+1)^4} + \dots \leq \dfrac{1}{6(n+1)^2n}.$

This implies $$ (n+1)^2 \left( \dfrac{a_{n+1}}{a_n} - 1 \right) \to 1/2$$ so upon multiplying the above by $na_n/(n+1)$ $$ n(n+1)(a_{n+1} - a_n) \to e/2 > 1.$$ Hence, $a_{n+1} - a_n \geq 1/n(n+1)$ for all large n.