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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...

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    RHS should have $2i$ , not $i$. – 2017-01-10
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    "The without induction" proof just implicitly uses induction, @S.C.B. – 2017-01-10
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    It's false for $n=1$. – 2017-01-10
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    changed it. thanks. – 2017-01-10

2 Answers 2

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

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