How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$
 A: Apply $\exp$ (which is strictly increasing) to both sides. Then your inequality is equivalent to
$$ \frac{n+1}{n}\le e^{\frac2{n+1}}$$
One of the most useful inequalities about the exponential is
$$ e^t\ge 1+t.$$
This gives us
$$ e^{\frac2{n+1}}\ge 1+\frac2{n+1}\stackrel{(*)}\ge1+\frac1n=\frac{n+1}n$$
where $(*)$ follows from $\frac2{n+1}-\frac1n=\frac{2n-(n+1)}{n(n+1)}\ge0$.
A: $$\ln((n+1)/n) = \ln(n+1)-\ln(n) = \int_n^{n+1}\frac{1}{x} dx \le \frac{1}{n} \le \frac{2}{n+1}$$ 
A: Take $x=\frac1n$ so you will have 
$$\ln(1+x)\leq \frac{2x}{x+1}$$ now suppose $f(x)=\ln(1+x)- \frac{2x}{x+1}\\x \in [0,1]$
$$f'(x)=\frac{1}{x+1}-\frac{2}{(x+1)^2}=\frac{1+x-2}{(x+1)^2} \to x=1$$ so $f'(x) \leq 0 \text {  when x is } \in [0,1]$ so $f(x)$ is decreasing in this interval 
so $$f(x)\leq f(0) \text{ when } x\in [0,1]\\ \to f(x)\leq f(0)\\f(x) \leq \ln(1+0)-0\\f(x)\leq 0\\\ln(1+x)- \frac{2x}{x+1}\leq 0\\
\ln(1+x)\leq \frac{2x}{x+1} , x \in [0,1] \\and \\\ln(1+x)\leq \frac{2x}{x+1} , x \in (0,1]$$ now put $x=\frac1n$
A: HINT 
Use the fact that $(1 + \frac 1 n)^n \lt e$ See Euler's constant
A: Start from: 
1) ln$ (1 +x) \le x$, for $x \ge 0$.
2) $ n +1 \le 2n$, 
$\rightarrow 1/n \le 2/(n+1)$.
Combining :  
ln$ (1+ 1/n) \le 1/n \le 2/(n +1)$, 
where  $n$ is  a positive integer.
Proof of 1):
Series expansion for $e^x$: 
$ e^x = 1+ x + x^2/2! + x^3/3! +....  $
$\rightarrow$ $(1+x) \le e^x$ for $x \ge 0$.
ln is a stricly monotonously increasing function , hence
ln $(1+ x) \le x$ for $x\ge0$.
A: Integrate $\frac{1}{x}$ between $n$ and $n+1$: $\ln \frac{n+1}{n} = \ln (n+1) - \ln (n) = \displaystyle\int_n^{n+1} \frac{1}{x} \operatorname{d}x$. As $x\to \frac{1}{x}$ is decreasing, you get $\ln \frac{n+1}{n}\leq \displaystyle\int_n^{n+1} \frac{1}{n} \operatorname{d}x = \frac{1}{n}$.
It is moreover clear that $\frac{1}{n} \leq \frac{2}{n+1}$ for $n\geq 1$, so you get $\ln \frac{n+1}{n} \leq \frac{2}{n+1}$
A: Your question is equivalent to prove the following form
$$
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \tag{1}
$$
The relation $(1)$ is true becuse $\frac{n-1}{n} \geq \frac{n-1}{n+1}$. 
Edit:
If the relation $(1)$ be true then we can conclude that:
$$
\left\{
\begin{array}{c}
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \\
\\
2^{(1-\frac{n-1}{n+1})}  \leq e^{(1-\frac{n-1}{n+1})} 
\end{array}
\right.
\Rightarrow 
(2-\frac{n-1}{n})\leq e^{(1-\frac{n-1}{n+1})} \tag{2}
$$
$$
\ln (2-\frac{n-1}{n})\leq (1-\frac{n-1}{n+1})
\Rightarrow  
\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1}
$$
