- #1
opticaltempest
- 135
- 0
I am given the following series and its sum
<br /> \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8}<br />
I need to find the sum of this series
<br /> \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} <br />
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,
<br /> a_n = \frac{1}{{\left( {2n - 1} \right)^2 }}<br />
<br /> = {\rm{\{ 1, 1/9, 1/25, }}...{\rm{\} }}<br />
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,
<br /> \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8} - 1 - \frac{1}{9}<br />
<br /> \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8}<br />
I need to find the sum of this series
<br /> \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} <br />
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,
<br /> a_n = \frac{1}{{\left( {2n - 1} \right)^2 }}<br />
<br /> = {\rm{\{ 1, 1/9, 1/25, }}...{\rm{\} }}<br />
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,
<br /> \sum\limits_{n = 3}^\infty {\frac{1}{{\left( {2n - 1} \right)^2 }}} = \frac{{\pi ^2 }}{8} - 1 - \frac{1}{9}<br />