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

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.

• Yes it is an inductive proof and it looks correct to me. – sudeep5221 Apr 16 at 1:53
• is not strictly neccessary get the some side(RHS or LHS) ? in this case, i develop a inequality seperate and after incorporate that proof to the original proof. Is this still valid for formal induction proof? – Eduardo Sebastián C. Apr 16 at 1:56
• It is just the way you have written the proof. It could have equivalently been written as $\displaystyle 2 - \frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} + \frac{1}{m+1} - \frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} - \frac{1}{m(m+1)} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} + \frac{1}{(m+1)}\left(\frac{1}{m+1} - \frac{1}{m} \right) \leq 2 - \frac{1}{m+1}$ – sudeep5221 Apr 16 at 2:00

## 2 Answers

Let me show you a different (more direct) way to prove this inequality. Note that $$k(k-1) and therefore $$\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 :)

• For more closer approximation of the series, use the inequality $k^2-1<k^2$ and repeat this same procedure with the partial fraction decomposition $$\dfrac{1}{k^2-1}=\dfrac1{2(k-1)}-\dfrac1{2(k+1)}.$$ – Bumblebee Apr 16 at 3:33

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

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