Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this?  I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one

$S = 1 +  \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +...  $ 
Take 1/4 common from the terms whose denominator is even.
$S = 1 +  \frac{1}{9} + \frac{1}{25} + ... +  \frac{1}{4} ( 1 + \frac{1}{4} +  \frac{1}{9} + ... ) $  
$S = 1 +  \frac{1}{9} + \frac{1}{25} + ... +  \frac{1}{4} S$  
$ 1 +  \frac{1}{9} + \frac{1}{25} + ...=    \frac{3}{4} S = (3/4)(\pi^2 /6) $  
