Sum of reciprocals of squares - bounding

Question

Recently in class our teacher told us about the evaluating of the sum of reciprocals of squares, that is $\sum_{n=1}^{\infty}\frac{1}{n^2}$. We began with proving that $\sum_{n=1}^{\infty}\frac{1}{n^2}<2$ by induction. However, we actually proved a stronger result, namely that $\sum_{n=1}^{\infty}\frac{1}{n^2}<2-\frac{1}{n}$, because we were told that it is much easier to prove the stronger result than the weaker one. My question is: why exactly is it easier? Is it because our induction hypothesis is stronger? And how to prove specifically the weaker result using induction, without actually evaluating the sum to be $\frac{\pi^2}{6}$?

Answer 1

To arrive at the result without induction, we note that (See this for a proof) an upper bound for the sum is given by

$$\begin{align} \sum_{n=1}^N\frac{1}{n^2}&\le 1+\int_1^N\frac{1}{x^2}\,dx\\\\ &=2-\frac1N \end{align}$$

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Now, if we proceed using induction, we first establish a base case. It is easy to see that for $N=1$

$$\sum_{n=1}^1 \frac1{n^2}=1\le 2-\frac11=1$$

Then, we assume that for some $N$ the inequality

$$\sum_{n=1}^N\frac1{n^2}\le 2-\frac1N$$

holds. Then, we see that

$$\begin{align} \sum_{n=1}^{N+1}\frac1{n^2}&=\sum_{n=1}^N\frac1{n^2}+\frac{1}{(N+1)^2}\\\\ &\le 2-\frac{1}{N}+\frac{1}{(N+1)^2}\\\\ &= 2-\frac{1}{N+1}+\left(\frac{1}{N+1}-\frac{1}{N}+\frac{1}{(N+1)^2}\right)\\\\ &= 2-\frac{1}{N+1}-\frac{1}{N(N+1)^2}\\\\ &\le 2-\frac{1}{N+1} \end{align}$$

And we are done!

Answer 2

You may Euler's acceleration method or just an iterated trick like my $(1)$ here to get: $$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = \color{red}{\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}}\tag{A}$$ and the last series converges pretty fast. Then you may notice that the last series comes out from a squared arcsine. Oh, wait, that just gives another proof of: $$ \zeta(2)=\frac{\pi^2}{6}. \tag{B}$$