# Sum the series

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#### chisigma

##### Well-known member
Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is...

$\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$

#### anil86

##### New member
Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is...

$\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is...

$\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$
Is...

$\displaystyle 1 - e^{2\ z} = - e^{z}\ (e^{z} - e^{- z}) -> \sqrt{1 - e^{2\ z}} = i\ e^{\frac{z}{2}}\ \sqrt {e^{z} - e^{- z}}\ (1)$

... so that for $\displaystyle z = i\ \theta$ is...

$\displaystyle \text{Im}\ \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\} = \text{Im}\ \{ \frac{e^{i\ \frac{\theta}{2}}}{i\ \sqrt{e^{i\ \theta} + e^{- i\ \theta}}} \} = \frac{\sin \frac{\theta}{2} + \cos {\frac{\theta}{2}}}{2\ \sqrt{\sin \theta}}\ (2)$

Kind regards

$\chi$ $\sigma$