Is the sum of this series correct?

[opticaltempest]

I am given the following series and its sum
$$\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8}$$
I need to find the sum of this series
$$\sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}}$$

Is the method I used below this the correct approach? Is there a better or different way?

From the first series where n=1 we have,

$$a_n = \frac{1}{{\left( {2n - 1} \right)^2 }}$$

$$= {\rm{\{ 1, 1/9, 1/25, }}...{\rm{\} }}$$

Therefore I concluded that we can subtract off the first two terms of
the above sequence to get the sum when n starts at 3. Hence,

$$\sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8} - 1 - \frac{1}{9}$$

 

[Wooh]

Your reasoning is correct.