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 $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
 A: 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 $x\geq 8$:
$$ f'(x)\geq \frac{\left(1+\frac{1}{8}\right)^8}{2(x+1)^2}-\frac{1}{x^2}>0.\tag{4} $$
A: 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.
A: 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 $n$.
A: Here is another way to approach this problem.
The function $$f(z) = 1 - z/2 + z^2/3 + \ldots + (-1)^{k+1} z^k/(k+1) + \ldots $$ is analytic on the unit disc $ \{ z : |z| < 1\}$, which implies $ g(z) = \exp f(z)$ is also analytic on $ \{ z : |z| < 1\}$ and hence can be expanded as a power series $$g(z) = a_0 + a_1 z + a_2 z^2 + \dots + $$ in $ \{ z : |z| < 1\}$. We can easily compute the the first few coefficients as $a_0 = \exp f(0) = e$, $a_1 = \exp f(0) f^{'}(0) = -e/2$, and similarly $a_3 = 11e/24.$
However, $f(x) = \log(1+x)/x$ for for all real $x$ with $ 0 < x < 1$, so $g(x) = (1+x)^{1/x}$ for $ 0 < x < 1$ and the above series for real $x$ is an analytic extension of $(1+x)^{1/x}$ to $-1 < x < 1$.
Writing $$(1+x)^{1/x} = e - ex/2 + 11x^2/24 + \dots  $$ from which we get
$(1+x)^{1/x} + cx = e + (c - e/2) x + 11ex^2/24 + \dots + \dots $.
The derivative of the above function at 0 is c -e/2, which is < 0, if c < e/2, by the continuity of the derivative, there is an interval $[0,\epsilon]$ on which the derivative of the function above is strictly negative and hence it decreases. Since 1/n decreases and lies in $[0,\epsilon]$ for all large $n$ this means $(1+1/n)^{n} + c/n$ increases eventually for any $ c < e/2$. This holds for any $x_n$ that strictly decreases to 0 not only $1/n,$, $(1+x_n)^{1/x_n} + c x_n$ eventually increases if $ c < e/2$. We can similarly argue that $(1+x_n)^{1/x_n} + c x_n$ eventually increases if $ c > e/2$ if $x_n$ strictly decreases to 0. For $c = e/2$, the positivity of the coefficient of $x^2$ implies $(1+x_n)^{1/x_n} + e x_n / 2$ eventually starts decreasing.
