Proof for $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$ I'm looking for a proof of $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$, optionally $\ln(n)+\frac{1}{n+1}\leq \ln(n+1)$
 A: If $x \ge 0$, 
$$e^x = 1 + \int_0^x e^t\, dt \le 1 + \int_0^x e^x\, dt = 1 + xe^x.$$
So $(1 - x)e^x \le 1$, and thus 
$$e^x \le \frac{1}{1 - x},\quad 0 \le x < 1.$$
Letting $x = \frac{1}{n+1}$, we obtain
$$e^{\frac{1}{n+1}} \le \frac{1}{1 - \frac{1}{n+1}} = \frac{n+1}{(n+1)-1} = \frac{n+1}{n} = 1 + \frac{1}{n}.$$
Equivalently, 
$$e^{\frac{1}{n+1}} - 1 - \frac{1}{n} \le 0.$$
A: Since $\ln(n+1)-\ln n$ represents the area under $y=\frac{1}{x}$ from $n$ to $n+1$ and $y=\frac{1}{x}$ is decreasing, 
we get that $\ln(n+1)-\ln n>\frac{1}{n+1}$.

Alternatively, applying the Mean Value Theorem to $f(x)=\ln x$ on $[n,n+1]$ gives
$\ln(n+1)-\ln n=f(n+1)-f(n)=f^{\prime}(c)=\frac{1}{c}>\frac{1}{n+1}$ for some $c$ in $(n,n+1)$.
A: In THIS ANSWER and THIS ONE, I showed using only (i) the limit definition of the exponential function and (ii) Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\frac{1}{1-x}\ge e^x\ge 1+x \tag 1$$
for $x<1$.  Letting $x=1/(n+1)$ in the left-hand inequality in $(1)$ yields
$$e^{1/(n+1)}\le \frac{1}{1-\frac{1}{n+1}}=1+\frac1n \tag 2$$
Rearranging $(2)$, we obtain the desired inequality
$$\bbox[5px,border:2px solid #C0A000]{e^{1/(n+1)}-1-\frac1n\le 0}$$ 
And we are done!
for 
A: for
$0 < x < 1$,
$e^x
=\sum_{n=0}^{\infty} \frac{x^n}{n!}
\lt\sum_{n=0}^{\infty} x^n
=\frac1{1-x}
$.
Therefore
$e^{1/(n+1)}
<\frac1{1-1/(n+1)}
=\frac{n+1}{n}
=1+\frac1{n}
$.
A: In this answer, it is shown that
$$
\left(1+\frac1n\right)^{n+1}
$$
is a decreasing sequence. Since $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{n+1}$, we have
$$
e\lt\left(1+\frac1n\right)^{n+1}
$$
Taking $n+1^{\text{st}}$ roots and subtracting yields
$$
e^{\large\frac1{n+1}}-1-\frac1n\lt0
$$
