Induction proof without summation I have to prove this induction:
$\dfrac{1}{(n+1)}+\dfrac{1}{(n+2)}+\dots+\dfrac{1}{2n} = \dfrac{1}{(1\times2)}+\dfrac{1}{(3\times4)}+\dots+\dfrac{1}{(2n-1)\times2n}$
Can someone help me with it?
 A: Hint. You may observe that
$$
\begin{align}
\left(\frac1{n+2}+\cdots+\frac1{2(n+1)}\right)-\left(\frac1{n+1}+\cdots+\frac1{2n}\right)&=-\frac1{n+1}+\frac1{2n+1}+\frac1{2(n+1)}\\\\
&=-\frac1{2(n+1)}+\frac1{2n+1}\\\\
&=\frac1{(2n+1)\times(2n+2)}.
\end{align}
$$
A: If you want to use the summation symbol, note that
$$
a_n=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=
\sum_{k=1}^n\frac{1}{n+k}
$$
Therefore
$$
a_{n+1}=\sum_{k=1}^{n+1}\frac{1}{n+1+k}
$$
The right-hand side can be written
$$
b_n=\sum_{k=1}^{n}\frac{1}{2k(2k-1)}
$$
So your task is to prove $a_n=b_n$. The case $n=1$ is trivial. Suppose the assertion holds for $n$. Then
\begin{align}
a_{n+1}&=\sum_{k=1}^{n+1}\frac{1}{n+1+k}\\[6px]
{\scriptsize\text{(add and subtract, detach two terms)}\quad}
&=-\frac{1}{n+1}+\biggl(\,\sum_{k=0}^{n-1}\frac{1}{n+1+k}\biggr)+
\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(set $l=k+1$)}\quad}
&=-\frac{1}{n+1}+\biggl(\,\sum_{l=1}^{n}\frac{1}{n+l}\biggr)+
\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(induction hypothesis)}\quad}
&=-\frac{1}{n+1}+b_n+\frac{1}{2n+1}+\frac{1}{2n+2}\\[6px]
{\scriptsize\text{(rearrange)}\quad}
&=b_n+\frac{1}{2n+1}-\frac{1}{2n+2}\\[6px]
&=b_n+\frac{2n+2-2n-1}{(2n+1)(2n+2)}\\[6px]
&=b_n+\frac{1}{(2n+1)(2n+2)}\\[6px]
&=b_{n+1}
\end{align}
