Proof inequality and series using Induction

Question

How can I prove that

$$ \sum_{k=1}^n \frac{1}{k^2} ≤ 2 - \frac {1}{n} $$

using induction?

I first tried to prove this for $P(1)$, so:

$$ \sum_{k=1}^n \frac{1}{1^2} ≤ 2 - \frac {1}1 $$ is proved. I then assume that $P(n)$ is proved.

Now I have to prove the same inequality for $P(n+1)$

$$ \sum_{k=1}^{n+1} \frac{1}{k^2} ≤ 2 - \frac {1}{n + 1} $$

I concentrate on the argument on the left and make some changes:

$$ \sum_{k=1}^{n+1} \frac{1}{k^2} $$

$$ \sum_{k=1}^{n} \frac{1}{k^2} + \frac {1}{n+1} $$

And since $P(n)$ is proved I can change it like this

$$ 2 - \frac {1}n + \frac {1}{n+1} $$

But I can't prove it, cause in the end I get a wrong inequality.

Someone can please tell me what I'm doing wrong?

Answer 1

Note that $$\sum_{k=1}^{n+1}\frac{1}{k^2}=\frac{1}{(n+1)^{\color{red}{2}}}+\sum_{k=1}^{n}\frac{1}{k^2}$$

Using the inductive hypothesis, we have $$\sum_{k=1}^{n+1}\frac{1}{k^2}=\frac{1}{(n+1)^2}+\sum_{k=1}^{n}\frac{1}{k^2}\le\frac{1}{(n+1)^2}+2-\frac 1n$$

Now, what you have to prove is $$\frac{1}{(n+1)^2}+2-\frac 1n\le 2-\frac{1}{n+1}$$

Answer 2

In the end I got that to complete the induction I only had to show $$1/n>(n+2)/(n+1)^2\Leftarrow n(n+2)<(n+1)^2$$ Which is obvious...

Answer 3

This is a non-induction solution but included for reference only.

$$\sum_{k=1}^n\frac 1{k^2}\leq 1+\int_1^n \frac 1{k^2}=1+\left[-\frac 1k\right]_1^n=2-\frac 1n$$