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TO BE PROVED: $$ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots \frac{1}{n^2}<2 $$
Please prove this by mathematical induction only. My approach: I already proved it through graphs but i have to prove it through mathematical induction. Prove this equation by mathematical induction that $\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + .... + \frac{1}{n^2} < 2 $ ?

Harshit Nyati
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  • You need to prove something like $\displaystyle\sum_{i=1}^n {1\over i^2} < 2-{1\over n}$ instead; the statement $$\sum_{i=1}^n {1\over i^2}<2$$ (which is what the link is to) doesn't give a strong enough induction hypothesis for induction to work. – Christopher Carl Heckman Feb 11 '16 at 06:52
  • Please write down the question instead of providing a link to a picture file. – Babai Feb 11 '16 at 06:55
  • i have wrote the question. please answer it now. – Harshit Nyati Feb 11 '16 at 07:00
  • @carl heckmen, sir please give a detailed proof. – Harshit Nyati Feb 11 '16 at 07:01
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    @Galc127 Next time, give the poster a chance to answer. Also, I wasn't sure that $\displaystyle 2-{1\over n}$ was the form that worked; I just answered off the top of my head. – Christopher Carl Heckman Feb 11 '16 at 07:04
  • For a non-induction but more straightforward approach: $$\sum_1^n\frac1{i^2}<1+\frac11\cdot\frac12+\frac12\cdot\frac13+\cdots+\frac1{n-1}\cdot\frac1n$$ then use $$\frac1{k-1}\cdot\frac1k=\frac1{k-1}-\frac1k$$ – Vim Feb 11 '16 at 07:26

1 Answers1

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First, let's make the claim stronger - $\displaystyle \forall n\in\Bbb N:\ 1+\frac{1}{2^2}+\frac{1}{3^2}+\dots+\frac{1}{n^2}\le2-\frac{1}{n}<2$

Base case - $n=1$ is trivial.

Step: assume the claim holds for $k$ and prove for $k+1$ $$\underbrace{1+\frac{1}{2^2}+\dots+\frac{1}{k^2}}_{\le 2-\frac{1}{k}\text{ by induction assumption}}+\frac{1}{(k+1)^2}\le 2-\frac{1}{k}+\frac{1}{(k+1)^2}\le 2-\frac{1}{k}+\frac{1}{k(k+1)}\\=2-\frac{1}{k}+\frac{1}{k}-\frac{1}{k+1}=2-\frac{1}{k+1}<2$$

Q.E.D

Galc127
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