<s>[INST] 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?
 [/INST] 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.
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