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

Question

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} $

Answer 1

Here is one way...

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

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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 $$ \sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2 $$