Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0$
This is not the starting inequality. 
Is there something wrong in this method?
 A: HINT: 
Take $$f(x)=\ln x $$ and apply MVT in [$a,a+1$].
A: For the second inequality multiply by $a$ to get 
$$\ln \left(1 + \frac 1a\right)^a < 1$$ or
$$\left(1 + \frac 1a\right)^a < e $$
which is correct since function $f(a) = \left(1 + \frac 1a\right)^a$ is strictly increasing and ha a limit at $+\infty$ equal to $e$.
To prove the first part use the function $g(a) = \left(1 + \frac 1a\right)^{a+1}$ which converges to $e$ from above.
A: METHOD 1: Non-Calculus Based
In This Answer, I showed using basic tools only that the logarithm function satisfies the inequalities 
$$\frac{x}{x+1}\le \log(1+x) \le x \tag 1$$
for $x\ge -1$.  Note that $\frac{a+1}{a}=1+\frac1a$.  Then, setting $x=\frac1a$ in $(1)$ gives the inequalities 
$$\frac{1}{a+1}\le \log \left(\frac{a+1}{a}\right)\le \frac1a$$
The strict inequalities follows since the equality in $(1)$ occurs only when $x=0$.  Since $\frac1a>0$, we have
$$\frac{1}{a+1} < \log \left(\frac{a+1}{a}\right) < \frac1a$$

METHOD 2: Calculus Based
Form the functions $f(x)=\log \left(1+x\right)-x$ and $g(x)=
\log(1+x)-\frac{x}{x+1}$ for $0<x$.  
Then, note that $f'(x)=\frac{-1}{1+x}<0$ and $g'(x)= \frac{x}{(x+1)^2}>0$.  
Therefore, $f(0)=0$ and $f'(x)<0$ implies $f(x) > 0$ for $x>0$.  Then set $x=1/a$ and we have 
$$\log \left(\frac{a+1}{a}\right) < \frac1a$$
Finally, $g(0)=0$ and $g'(x)>0$ implies $g(x)>0$.  Then, set $x=1/a$ and we have
$$\frac{1}{a+1}< \log \left(\frac{a+1}{a}\right) $$
And we are done.
A: Hint: use $f(x)=\ln x$ in $[a,a+1]$.
