Proving sum of $1/n^2$ is less than or equal to $2$ So I'm suppose to prove that $\sum 1/n^2 \le 2$. Should I use induction?
 A: Hint: Prove the the following holds for all $n$ by induction. 
$$\sum_1^n \frac{1}{k^2} \le 2 - \frac{1}{n}.$$
Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work. 
You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$. 
A: If you don't want to use induction, you can use the integral test for convergence which says that for a non-negative, strictly decreasing function $f$, we have $$\int_N^{\infty}f(x)dx \leq \sum_{n=N}^{\infty}f(n) \leq f(N) + \int_N^{\infty}f(x)dx.$$ In this case $f(x) = \dfrac{1}{x^2}$ and $N = 1$.
A: 
Should I use induction?

Not necessarily. Note that $\dfrac1{n^2}\lt\dfrac1{n(n-1)}=\dfrac1{n-1}-\dfrac1n$ for each $n\geqslant2$ hence, telescoping the terms of the series for $n\geqslant2$ shows that
$$
\sum_{n=2}^{\infty}\frac1{n^2}\lt\sum_{n=\color{red}{\mathbf2}}^{\infty}\left(\dfrac1{n-1}-\dfrac1n\right)=\dfrac1{\color{red}{\mathbf2}-1}=1.
$$
To conclude, add the $n=1$ term $\dfrac1{1^2}=1$.
A: In potato's answer,it may seem that the stronger induction hypothesis came out of the blue. Here is another approach:
Hint: For all $n\in\mathbb{Z}^+$, we have: $\frac{1}{n^2}\leq\frac{1}{n^2-0.5^2}=\frac{1}{(n-0.5)(n+0.5)}$
Thus:
$$\sum_{n=1}^\infty \frac{1}{n^2}\leq \sum_{n=1}^\infty \frac{1}{(n-0.5)(n+0.5)}$$
The sum in the right hand side of the inequality telescopes and can be shown to be equal to $2$.
A: There's a geometric proof that the sum of $1/n$ is less than 2.
One divides a square into rows of height 1/2, 1/4, 1/8, 1/16 &c.
Onto the top shelf of height 1/2, go 1/2, 1/3.
Onto the second shelf (1/4), go 1/4, 1/5, 1/6, 1/7, all equal to or less or equal to 1/4, so they will all fit.  
And so on with 1/8, 1/16, &c.  
Since all the squares $1/2^2$ to $1/\infty^2$ fit into the second square, then the sum of these numbers must be less than $2$.
