Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction?

Question

I have this statement:

Prove by induction that $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ for $n \in \mathbb{N}$

My attempt was:

Base case: $\frac{1}{1} \leq 1$

Assume that $\displaystyle \sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$ is true for some $m \in \mathbb{N}$

To prove: $\displaystyle \sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$

Since $$\sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$$

$$\sum_{k = 1}^m\frac{1}{k^2} + \frac{1}{(m+1)^2}\leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$

$$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$

Now, I'll prove that (1) $$\displaystyle 2 -\frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m + 1}$$

Developing this inequation, gives $$m^2 + 2m + 1 \geq m^2 + 2m$$ getting that (1) is true.

And by transitivity $$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$$ was proved.

But I don't know if this is a valid induction proof and if not, what is the type of this proof?

Thanks in advance.

Answer 1

Let me show you a different (more direct) way to prove this inequality. Note that $k(k-1)<k^2$ and therefore $$\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}$$ for all $k>1.$
Now for $n>1$ we have $$\sum_{k=1}^n\dfrac{1}{k^2}\le1+\sum_{k=2}^n\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)=1+\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\cdots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right).$$ Simplify the RHS and see it for your self :)

Answer 2

If we know that it is valid for $n$, we only need to show that $\frac{1}{(n+1)^{2}}<\frac{1}{n}-\frac{1}{n+1}$

Therefore

$$ \sum_{k=1}^{n}{\frac{1}{k^{2}}}+\frac{1}{(n+1)^{2}}<2-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1} $$