<s>[INST] 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
 [/INST] EDIT: This approach is unfortunately wrong, as pointed out in the comments.
Let $S(n) = \frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}.$
(1) Base step. Let $n=1$. Then $n+1 = 2n$, so the series only has one term. $S(1) = \frac{1}{2} < \frac{25}{36}$, hence the statement is true for $n=1$.
(2) Inductive step. Suppose the induction hypothesis is true. Then
$S(n+1) = \frac{1}{ (n+1)+1}+\frac{1}{ (n+1)+2}+\cdots+\frac{1}{2(n+1)}$
$= \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}$
$= - \frac{1}{n+1} + \left(\frac{1}{n+1} + \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}\right) + \frac{1}{2n+1} + \frac{1}{2n+2}$
$= - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + S(n)$
$< - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + \frac{25}{36}$
where the inequality comes from the induction hypothesis.
Now $S(n+1) < \frac{25}{36}$, as we want to show, requires
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < 0.$
As $- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1}$, this is certainly true if
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1} < 0$, or $\frac{1}{n+1} < \frac{2}{2n+1}$. The last inequality is trivial to show.
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