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

I didn't prove it by mathematical induction.
Set $a_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$, then by calculation, $a_{n+1}-a_{n}=\frac{1}{2n+1}-\frac{1}{2n+2}$, $a_{n}$ is strictly monotonically increasing.
So we can obtain that
$$\begin{align*}
a_{n}&=a_{1}+\sum_{k=1}^{n-1}(a_{k+1}-a_{k})\\
&=\frac{1}{2}+\sum_{k=1}^{n-1}(\frac{1}{2k+1}-\frac{1}{2k+2})\\
&=\sum_{k=0}^{n-1}(\frac{1}{2k+1}-\frac{1}{2k+2})\\
&=\sum_{k=0}^{n-1}\int_{0}^{1}(x^{2k}-x^{2k+1})dx\\
&=\int_{0}^{1}\sum_{k=0}^{n-1}(x^{2k}-x^{2k+1})dx\\
&=\int_{0}^{1}\frac{1-x^{2n}}{1+x}dx\\
&=\ln2-\int_{0}^{1}\frac{x^{2n}}{1+x}dx
\end{align*}$$
When $x\in[0,1]$, $\frac{x^{2n}}{2}\leq\frac{x^{2n}}{1+x}\leq x^{2n}$, integraling at both sides, we can get that $(0<)\frac{1}{2(2n+1)}\leq\int_{0}^{1}\frac{x^{2n}}{1+x}dx\leq\frac{1}{2n+1}$, so for arbitrary $n$, $a_{n}<\ln2<\frac{25}{36}$.
