Find the$A:=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{20^2}$ 
Find the :
$$A:=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{20^2}$$


My Try :
$$\frac{1}{2^2}+\frac{1}{3^2}=\frac{3^2+2^2}{2^2\cdot 3^2}$$
$$\frac{3^2+2^2}{2^2\cdot 3^2}+\frac{1}{4^2}=\frac{(3^2+2^2)(4^2+1)}{4^2\cdot 3^2\cdot 2^2}$$
$$\frac{(3^2+2^2)(4^2+1)}{4^2\cdot 3^2\cdot 2^2}+\frac{1}{5^2}=\frac{(5^2)(3^2+2^2)(4^2+1)+(4^2)(3^2)(2^2)}{5^2 \cdot4^2\cdot3^2\cdot 2^2}$$
Now what ?

We know that the sum of the reciprocal of the squares add up to $\pi^2/6.$  So your sum equals
$$\frac{\pi^2}{6} -1 -\sum_{n=21}^{\infty} \frac{1}{n^2}.$$
The infinite sum is less than, but very close to
$$\int_{20}^\infty \frac{1}{x^2} \; dx = \frac{1}{20}.$$
So your sum is about 
$$\frac{\pi^2}{6} -1 -\frac{1}{20} = 0.594934068\ldots.$$
Maple says the exact answer is $0.5961632439.$
