$\forall n \ge 1, \sum _{i=1}^n \frac 1 {i^2} \le 2$ mathematical induction

Question

Prove using mathematical induction:

$$\forall n \ge 1, \sum _{i=1}^n \frac 1 {i^2} \le 2$$

I have seen this problem from the note of CS70 which is Berkeley's discrete math course. This problem illustrates strengthening the induction hypothesis. First, we can not prove that using induction, but then the note tells that we can prove that is less than $2-\frac 1 n$, so my question is how can we think about the question from proving it less than 2 to less than $2- \frac 1 n$? Thanks for your help!

Answer 1

Since $a_n=\sum_{k=1}^{n}\frac{1}{k^2}$ is obviously an increasing sequence (I prefer to avoid the letter $i$ as a summation index, since it can be easily mistaken with the imaginary unit) it is enough to show that $\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}$ is less than two. Fourier theory ensures that $$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}$$ for any $x\in(0,2\pi)$, with such identity holding uniformly over compact subsets of $(0,2\pi)$. Since $$ \int_{0}^{2\pi}\sin(nx)\sin(mx)\,dx = \pi \delta(m,n)$$ we have $$ \int_{0}^{2\pi}\left(\frac{\pi-x}{2}\right)^2\,dx = \pi \zeta(2)$$ and the given claim is equivalent to $$ \int_{0}^{2\pi}(\pi -x)^2\,dx = 2\int_{0}^{\pi}(\pi -x)^2\,dx = 2\int_{0}^{\pi}x^2\,dx < 8\pi $$ or to $$ \pi < 2\sqrt{3} $$ which is equivalent to the perimeter of the regular hexagon circumscribed to a unit circle is greater than the perimeter of the unit circle. This follows from the fact that if $A,B$ are bounded convex sets in $\mathbb{R}^2$ with $B\subsetneq A$, then $\mu(\partial B)<\mu(\partial A)$.

Answer 2

Prove by induction that $\sum_{i=2}^n\frac{1}{i(i-1)}=1-\frac{1}{n}$ so $\sum_{i=1}^n\frac{1}{i^2}\le 2-\frac{1}{n}$.