prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
 $$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1

Here's an elementary proof, that requires no lengthy computations. Let 
$$S_n=\sum_{k=n+1}^{2n}\frac1k$$
We show by induction that $S_n \le \frac{25}{36} - \frac{1}{4n+1}$ for all $n \ge 2$. To start with, $S_2 = \frac13+\frac14=\frac{25}{36} - \frac19$, so the hypothesis is true for $n=2$. Now suppose it is true for $n-1$. Then
$$\begin{align}
S_n &=  S_{n-1} -\frac1n + \frac{1}{2n-1} + \frac{1}{2n}\\
&= S_{n-1} + \frac{1}{2n(2n-1)}\\
&\le \frac{25}{36} - \frac{1}{4n-3} + \frac{1}{2n(2n-1)}\\
&= \frac{25}{36} - \frac{1}{4n+1} - \frac{4}{(4n-3)(4n+1)} + \frac{1}{2n(2n-1)}\\
&< \frac{25}{36} - \frac{1}{4n+1}
\end{align}$$
because $2n(2n-1) > (4n-3)(4n+1)/4$.
