To prove this we will show that this sequence is bounded as well as monotonic
First , we prove that it is bounded . for this, we have
\left | s_n \right |=s_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}< \frac{1}{n}+...+\frac{1}{n}\;\; up\;to\;n\;terms=n.\frac{1}{n}=\Rightarrow \left | s_n \right |<1 \;\;for \;all\; nthus , the sequence <sn> is bounded . Now , we have to prove that it is monotonic
for this we have
s_{n+1}-s_n=\left ( \frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+2} \right )-\left ( \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \right )\\ \\ =\frac{1}{2n+1}-\frac{1}{2n+2}-\frac{1}{n+1}\\ \\ =\frac{1}{2n+1}-\frac{1}{2n+2}>0Thus the sequence <s_n> is monotonically increasing . since <s_n> is bounded and monotonically increasing sequence hence its converges.