Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ 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} $
 A: 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.
A: 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.
A: Here is one way...
$$\frac1{k^2} < \frac1{k(k-1)}= \frac1{k-1} - \frac1{k}$$
Now telescope to get
$$1+\sum_{k=2}^n\frac1{k^2} < 1+1-\frac1{n}< 2$$
A: $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}+...\uparrow\frac{\pi^2}{6}=1,6449340668482...\lt2.$$
See: http://mathworld.wolfram.com/PiFormulas.html
A: 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
$$
\sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2
$$
A: Find the fourier series of $f(x)=x^2$ on $(-1,1)$. It can give you the accurate solution which is $\dfrac {\pi ^2}{6}$.
