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Is the following Proof Correct?

Theorem. For $n\ge 1$, $$\sum_{j=1}^{n}\frac{1}{j^2}\leq 2-\frac{1}{n}$$ Proof. We construct the proof by recourse to Mathematical-Induction.

For $n=1$ we see that $\sum_{j=1}^{1}\frac{1}{1} = 1\leq \frac{3}{2} = 2-\frac{1}{2}$ establishing the basis of the induction.

Now let $k$ be arbitrary and assume for $k$ that $$\sum_{j=1}^{k}\frac{1}{j^2}\leq 2-\frac{1}{k}$$ consequently $$\sum_{j=1}^{k+1}\frac{1}{j^2} = \sum_{j=1}^{k}\frac{1}{j^2}+\frac{1}{(k+1)^2}\leq 2 - \frac{1}{k}+\frac{1}{(k+1)^2}$$ but observing the following equivalence $$2 - \frac{1}{k}+\frac{1}{(k+1)^2} = 2-\left(1+\frac{1}{k(k+1)}\right)\cdot\frac{1}{k+1}$$ together with the fact that $k\ge 1$ implies that $$\sum_{j=1}^{k+1}\frac{1}{j^2}\leq 2-\left(1+\frac{1}{k(k+1)}\right)\cdot\frac{1}{k+1}\leq 2-\frac{1}{k+1}$$ completing the inductive step.

$\blacksquare$

atifcppprogrammer
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1 Answers1

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Yes, you are right. Personally I would elaborate the equation $$-\frac1k+\frac1{(k+1)^2} = -\left(1+\frac1{k(k+1)}\right)\cdot\frac1{k+1}$$ It is correct but not directly obvious. I had to check it on a paper too...

But all in all your proof is good.

Mundron Schmidt
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