prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
 $$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}&lt;\frac{25}{36}$$

by Mathematical induction，n$\ge $1
As far as I can tell, none of the answers so far gives a proof by induction. Here's one.

Let

$$S_n=\sum_{k=n+1}^{2n}\frac1k\;.$$

By explicit computation

$$\frac{25}{36}-\frac1{60}-S_{16}=\frac{238793353}{20629078984800}\gt0\;.$$

As others have pointed out, the sequence monotonically increases, so this establishes the inequality up to $n=16$. Now we can prove

$$
S_n\lt\frac{25}{36}-\frac1{4(n-1)}
$$

for $n\ge16$ by induction. The base case is handled above. The induction step is

$$
\begin{align}
S_{n+1}
&amp;=
S_n+\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}
\\
&amp;=
S_n+\frac1{(2n+1)(2n+2)}
\\
&amp;\lt
S_n+\frac1{4n(n-1)}
\\
&amp;\lt
\frac{25}{36}-\frac1{4(n-1)}+\frac1{4n(n-1)}
\\
&amp;=
\frac{25}{36}-\frac1{4n}\;.
\end{align}
$$

The general insight to be gained here is that a proof by induction sometimes becomes easier when we make the conclusion stronger. That's because this not only strengthens the conclusion of the induction step but also its premise. In the present case, the original claim can't be proved directly by induction, since knowing that the inequality holds doesn't help in proving that it holds if we add something to the lesser side; but by strengthening the conclusion slightly more for $n$ than for $n+1$ we can create some space to work with in the induction step.
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. $S(1) = \frac{1}{2} &lt; \frac{25}{36}$, hence the statement is true for $n=1$.

(2) Inductive step. Suppose the induction hypothesis is true. Then

$S(n+1) = \frac{1}{ (n+1)+1}+\frac{1}{ (n+1)+2}+\cdots+\frac{1}{2(n+1)}$

$= \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}$

$= - \frac{1}{n+1} + \left(\frac{1}{n+1} + \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}\right) + \frac{1}{2n+1} + \frac{1}{2n+2}$

$= - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + S(n)$

$&lt; - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + \frac{25}{36}$

where the inequality comes from the induction hypothesis.

Now $S(n+1) &lt; \frac{25}{36}$, as we want to show, requires
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} &lt; 0.$

As $- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} &lt; - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1}$, this is certainly true if

$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1} &lt; 0$, or $\frac{1}{n+1} &lt; \frac{2}{2n+1}$. The last inequality is trivial to show.