Q:

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 $(n,n+1)$.

