Sum of reciprocals of squares - bounding 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 $\frac{\pi^2}{6}$?
 A: To arrive at the result without induction, we note that (See this for a proof) an upper bound for the sum is given by
$$\begin{align}
\sum_{n=1}^N\frac{1}{n^2}&\le 1+\int_1^N\frac{1}{x^2}\,dx\\\\
&=2-\frac1N
\end{align}$$

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!
A: 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:
$$ \zeta(2)=\frac{\pi^2}{6}. \tag{B}$$
