Induction proof without summation

Question

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?

Answer 1

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}

Answer 2

Hint. You may observe that $$ \begin{align} \left(\frac1{n+2}+\cdots+\frac1{2(n+1)}\right)-\left(\frac1{n+1}+\cdots+\frac1{2n}\right)&=-\frac1{n+1}+\frac1{2n+1}+\frac1{2(n+1)}\\\\ &=-\frac1{2(n+1)}+\frac1{2n+1}\\\\ &=\frac1{(2n+1)\times(2n+2)}. \end{align} $$