Mathematical induction for inequalities: $\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$ Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$$
adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is true by the inductive hypothesis, the 1 is already less than all those values? 

More generally
(one of my favorite phrases),
let
$s_k(n)
=\sum\limits_{i=n+1}^{kn+1} \frac1{i}
$.
I will show that
$s_k(n+1)>s_k(n)$
for $k \ge 3$.
In particular,
for $n \ge 1$
$s_3(n)
\ge s_3(1)
=\frac1{2}+\frac1{3}+\frac1{4}
=\frac{6+4+3}{12}
=\frac{13}{12}
> 1
$.
$\begin{array}\\
s_k(n+1)-s_k(n)
&=\sum\limits_{i=n+2}^{kn+k+1} \frac1{i}-\sum\limits_{i=n+1}^{kn+1} \frac1{i}\\
&=\sum\limits_{i=n+2}^{kn+1} \frac1{i}+\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i}
-\left(\frac1{n+1}+\sum\limits_{i=n+2}^{kn+1} \frac1{i}\right)\\
&=\sum\limits_{i=kn+2}^{kn+k+1} \frac1{i}-\frac1{n+1}\\
&=\sum\limits_{i=2}^{k+1} \frac1{kn+i}-\frac1{n+1}\\
&=\frac1{kn+2}+\frac1{kn+k+1}+\sum\limits_{i=3}^{k} \frac1{kn+i}-\frac1{n+1}\\
\end{array}
$
$\sum\limits_{i=3}^{k} \frac1{kn+i}
\ge \sum\limits_{i=3}^{k} \frac1{kn+k}
= \frac{k-2}{kn+k}
$.
If we can show that
$\frac1{kn+2}+\frac1{kn+k+1}
\ge \frac{2}{kn+k}
$,
then
$s_k(n+1)-s_k(n)
\ge \frac{2}{kn+k}+\frac{k-2}{kn+k}-\frac1{n+1}
= \frac{k}{kn+k}-\frac1{n+1}
= \frac{1}{n+1}-\frac1{n+1}
=0
$.
But
$\begin{array}\\
\frac1{kn+2}+\frac1{kn+k+1}-\frac{2}{kn+k}
&=\frac{kn+k+1+(kn+2)}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\
&=\frac{2kn+k+3}{(kn+2)(kn+k+1)}-\frac{2}{kn+k}\\
&=\frac{(2kn+k+3)(kn+k)-2(kn+2)(kn+k+1)}{(kn+2)(kn+k+1)(kn+k)}\\
\end{array}
$
Looking at the numerator,
$\begin{array}\\
(2kn+k+3)(kn+k)-2(kn+2)(kn+k+1)
&=2k^2n^2+kn(k+3+2k)+k(k+3)\\
&-2(k^2n^2+kn(k+3)+2(k+1)\\
&=2k^2n^2+kn(3k+3)+k(k+3)\\
&-2k^2n^2-2kn(k+3)-4(k+1)\\
&=kn(3k+3)+k(k+3)\\
&-kn(2k+6)-4(k+1)\\
&=kn(k-3)+k(k+3)-4(k+1)\\
&=kn(k-3)+k^2-k-4\\
&> 0 \quad\text{for $k \ge 3$}
\end{array}
$
and we are done.
