[tex]\text{Vom folosi inegalitatile:}\\
k^2\ \textless \ k^2+1\ \textless \ (k+1)^2,k\ \textgreater \ 1\\
\text{Prin urmare:}\\
k\ \textless \ \sqrt{k^2+1}\ \textless \ k+1\\
\displaystyle\sum_{k=1}^n \dfrac{1}{n+k+1}\ \textless \ \displaystyle\sum_{k=1}^n \dfrac{1}{n+\sqrt{k^2+1}}\ \textless \ \displaystyle\sum_{k=1}^n \dfrac{1}{n+k}\\
\dfrac{1}{n+2}+\dfrac{1}{n+3}+\ldots\+\dfrac{1}{2n+1}\ \textless \ \displaystyle\sum_{k=1}^n \dfrac{1}{n+\sqrt{k^2+1}}\ \textless \ \dfrac{1}{n+1}+\ldots+\dfrac{1}{2n}\\
\text{Trecand la limita obtinem:}\\
[/tex]
[tex]\ln 2\ \textless \ \displaystyle\limit\lim_{n\to \infty} \displaystyle\sum_{k=1}^n \dfrac{1}{n+\sqrt{k^2+1}}\ \textless \ \ln 2 \\
\text{Din teorema clestelui rezulta ca }\displaystyle\limit\lim_{n\to \infty} \displaystyle\sum_{k=1}^n \dfrac{1}{n+\sqrt{k^2+1}}=\boxed{\ln 2} [/tex]
