\xi\in(a+1,a) \\
f(a) - f(a+1) = f'(\xi)(a-a-1) \\
f(a+1) - f(a) = f'(\xi) \\

f(x) = \ln{x} \\
\\
f'(\xi) = {1\over(\xi)}

\ln{(a+1)\over a} = {1\over (\xi)}

(a+1)< \xi < a

\Rightarrow
\frac{1}{1+a}<\ln(1+{1 \over a})<\frac{1}{a}