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Base case: n=1. $1/1\le 2-1/1$. So the base case holds.

Let $n=k\ge1$ and assume

$$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$

We want to prove this for $k+1$, i.e.

$$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$

This is where I get stuck. Any help appreciated.

Fly by Night
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blutuu
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4 Answers4

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For your induction step, all you need is $$ 2-\frac{1}{k}+\frac{1}{(k+1)^2}\leq 2-\frac{1}{k+1}. $$ That's equivalent to $$ \frac{1}{k}-\frac{1}{(k+1)^2}-\frac{1}{k+1}\geq 0 $$ i.e. $$ \frac{(k+1)^2-k-k(k+1)}{k(k+1)^2}=\frac{1}{k(k+1)^2}\geq 0. $$ So it holds.

Julien
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  • Could you explain the first step for me. I'm confused about where the 2 came from on the LHS. – blutuu Apr 04 '13 at 15:42
  • @blutuu Induction hypothesis: $\sum_{j=1}^k1/j^2\leq 2-1/k$. Hence $\sum_{j=1}^{k+1}1/j^2\leq 2-1/k+ 1/(k+1)^2$ by adding $1/(k+1)^2$ on both sides of the previous inequality. So it suffices to prove $2-1/k+ 1/(k+1)^2\leq 2-1/(k+1).$ – Julien Apr 04 '13 at 15:45
  • So would the $2-1/k$ on the LHS basically come from $1/k^2<=2-1/k+1/(k+1)^2<=2-1/(k+1)$ omitting the first part of the inequality? – blutuu Apr 04 '13 at 16:03
  • @blutuu No. Read my comment more carefully and let me know if you still can't see why. You just add $1/(k+1)^2$ on both sides. Then it is the rhs that you want to make smaller than $2-1/(k+1)$. – Julien Apr 04 '13 at 16:09
  • @blutuu Actually, I guess we can say what you said in your previous comment. Except that the lhs of your two inequalities does not make sense. – Julien Apr 04 '13 at 16:13
  • I understand that. Thanks. – blutuu Apr 04 '13 at 16:13
4

$\displaystyle \frac{1}{i^2}< \frac{1}{i(i-1)}=\frac{1}{i-1}-\frac{1}{i}$,for $i\ge 2$

So we have $\displaystyle 1+\sum_{i=2}^{k}\frac{1}{i^2}\le 1+\sum_{i=2}^{k}(\frac{1}{i-1}-\frac{1}{i})=2-\frac{1}{k}$

I think this is better than induction.

Abhra Abir Kundu
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Let's assume it's true for $k$. Then $\sum_{i=1}^k\dfrac{1}{i^2}\leq2-\dfrac 1 k$

Let's try $k+1$.

$\sum_{i=1}^{k+1}\dfrac{1}{i^2}=\sum_{i=1}^k\dfrac{1}{i^2}+\dfrac{1}{(k+1)^2}\leq2-\dfrac 1 k+\dfrac{1}{(k+1)^2}$

As $\dfrac{1}{(k+1)^2}-\dfrac 1 k<-\dfrac{1}{k+1}$ ,

$\sum_{i=1}^{k+1}\dfrac{1}{i^2}\leq2-\dfrac{1}{(k+1)^2}$ which shows that it is true for $k+1$.

Ishan Banerjee
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0

I will give only a hint. Where you get stuck, you have by induction hypothesis that $$1/1^2+\ldots+1/k^2\leq 2-1/k$$ Subtituting this in $$1/1^2+\ldots+1/k^2+1/(k+1)^2$$ gives you $$1/1^2+\ldots+1/k^2+1/(k+1)^2\leq 2-1/k+1/(k+1)^2$$ And then, you have only to prove $$2-1/k+1/(k+1)^2\leq 2-1/(k+1)$$ Here, I think you will be able to rewrite the inequality in order to prove it.

Josué Tonelli-Cueto
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