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How many times more than $0$?

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

Let $n=k\ge1$ and assume

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

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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.

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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.

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

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 $k+1$.

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.

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