Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ [duplicate]

Having the following inequality

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$ To prove it for all natural numbers is it enough to show that:

$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2}$

marked as duplicate by Martin Sleziak, graydad, Rolf Hoyer, Erick Wong, Steven StadnickiApr 30 '15 at 15:41

• Is it required to use induction? – Vim Feb 16 '15 at 11:04
• no, diffrent methods are allowed but probably induction will be the easiest – kurkowski Feb 16 '15 at 11:16
• Probably induction is the worst method to be honest in that case... – Martigan Feb 16 '15 at 11:17
• does not matter could you show me any other valid method for this proof? – kurkowski Feb 16 '15 at 11:17
• The first inequality is obvious because the first term is negative and the second equality is not sufficient, you should see this post:math.stackexchange.com/questions/718872/… – Elaqqad Feb 16 '15 at 11:41

Here is one way...

$$\frac1{k^2} < \frac1{k(k-1)}= \frac1{k-1} - \frac1{k}$$

Now telescope to get $$1+\sum_{k=2}^n\frac1{k^2} < 1+1-\frac1{n}< 2$$

• That's the way to go. – Simon S Feb 16 '15 at 12:10

By using some telescopic sums, we can be even more accurate. We may notice that: $$\frac{1}{n^2}-\frac{1}{n(n-1)} = -\frac{1}{n^2(n-1)}$$ and if $n>1$: $$\frac{1}{n^2(n-1)}-\frac{1}{(n-1)n(n+1)}=\frac{1}{(n-1)n^2(n+1)}$$ so: $$\sum_{n=1}^{N}\frac{1}{n^2} = 1+\sum_{n=2}^{N}\frac{1}{n(n-1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n(n+1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}$$ or: $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{1}{n^2} &=& 2-\frac{1}{N}-\frac{N^2+N-2}{4N(N+1)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}\\&=&\frac{7}{4}-\frac{2N+1}{2N(1+N)}-\sum_{n=2}^{N}\frac{1}{(n-1)n^2(n+1)}\\&\leq&\color{red}{\frac{7}{4}}-\frac{1}{N+1}.\end{eqnarray*}$$

The telescopic approach (or the Euler's acceleration method) also leads to:

$$\forall N\geq 2,\qquad \sum_{n=1}^{N}\frac{3}{n^2\binom{2n}{n}}\leq \zeta(2) \leq \frac{1}{N^2\binom{2N}{N}}+\sum_{n=1}^{N}\frac{3}{n^2\binom{2n}{n}}$$

so, by just choosing $N=3$, $$\zeta(2)\leq\color{red}{\frac{593}{360}}$$ and the approximation is accurate up to two figures.

• Nice +1. Using $\dfrac1{n^2} < \dfrac1{n^2-1} = \dfrac12\left(\dfrac1{n-1}-\dfrac1{n+1} \right)$ to telescope itself gives $\dfrac74$. – Macavity Feb 16 '15 at 13:45

As noted, induction is a more difficult way to prove this. Here it is. Claim: $$\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}$$ for $n=2,3,4,\cdots$. First, when $n=2$ we have $$\frac{1}{1}+\frac{1}{4} = \frac{5}{4} < \frac{3}{2} = 2-\frac{1}{2}$$ which is correct.
Now, suppose $n \ge 2$ and $$\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}$$ Then $$\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}+\frac{1}{(n+1)^2} < 2-\frac{1}{n}+\frac{1}{(n+1)^2}$$ so it suffices to prove $$2-\frac{1}{n} + \frac{1}{(n+1)^2} < 2-\frac{1}{n+1}$$ This is equivalent to $$\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$$ which holds iff $$n+n(n+1) < (n+1)^2$$ or $$2n+n^2 < 1+2n+n^2$$ which is true. This completes the induction.

• Sometimes we get a question asking for a situation where proving something stronger makes the induction easier. Here is one of those! – GEdgar Feb 16 '15 at 14:18

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}+...\uparrow\frac{\pi^2}{6}=1,6449340668482...\lt2.$$ See: http://mathworld.wolfram.com/PiFormulas.html

The sum $\sum_{k=2}^n\frac{1}{k^2}$ is a lower Riemann sum for the function $1/x^2$ in the interval $[1,n]$, so $$\sum_{k=2}^n\frac{1}{k^2}\le\int_{1}^n\frac{1}{x^2}\,dx= \left[-\frac{1}{x}\right]_1^n=1-\frac{1}{n}$$ Therefore $$\sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2$$

Find the fourier series of $f(x)=x^2$ on $(-1,1)$. It can give you the accurate solution which is $\dfrac {\pi ^2}{6}$.