prove by induction expression

Question

Show that for all natural numbers $n$, the following equality holds:

$$\sum_{i=1}^{n}\frac{1}{2i(2i-1)} = \sum_{i=n+1}^{2n}\frac{1}{i}$$

Can't seem to wrap my head around it...

Answer 1

By induction hypothesis $$ \sum_{i=1}^{n+1}\frac{1}{2i(2i-1)}= \frac{1}{(2n+1)(2n+2)}+\sum_{i=n+1}^{2n}\frac{1}{i} $$ On the other hand $$ \sum_{i=n+2}^{2n+2}\frac{1}{i}= -\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}+\sum_{i=n+1}^{2n}\frac{1}{i} $$ So you just have to prove that $$ -\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\frac{1}{(2n+1)(2n+2)} $$

Answer 2

Hint: $\frac{1}{2i(2i-1)}=\frac{1}{2i-1}+\frac{1}{2i}-\frac{1}{i}$