Step by step solution to "Prove that:∑n=1∞sin⁡(nθ)n=π2−θ2"

fecundavai3c

Answered question

2022-02-27

Prove that:

\sum_{n = 1}^{\infty} \frac{\sin (n θ)}{n} = \frac{π}{2} - \frac{θ}{2}

besplodnexkj

Beginner2022-02-28Added 7 answers

Let,

S_{1} = \sum_{n = 1}^{\infty} \frac{\cos n θ}{n}

S_{2} = \sum_{n = 1}^{\infty} \frac{\sin n θ}{n}

Then

S_{1} + i S_{2} = \sum_{n = 1}^{\infty} \frac{\cos (n θ) + i \sin (n θ)}{n} = \sum_{n = 1}^{\infty} \frac{e^{\in θ}}{n}

Now, from the Taylor expansion,

\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} \dots

\Rightarrow - \ln (1 - x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} \dots = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

S_{1} + i S_{2} = - \ln (1 - e^{i θ})

= - \ln (1 - \cos θ - i \sin θ)

= - \ln (2 \frac{\sin^{2} θ}{2} - 2 i \sin (\frac{θ}{2}) \cos (\frac{θ}{2}))

= - \ln (2 \frac{\sin θ}{2}) - \ln (\frac{\sin θ}{2} - i \frac{\cos θ}{2})

= - \ln (2 \frac{\sin θ}{2}) + \ln (e^{i (\frac{π}{2} - \frac{θ}{2})})

= - \ln (2 \frac{\sin θ}{2}) + i (\frac{π}{2} - \frac{θ}{2})

Taking the imaginary part of both sides,

S_{2} = \frac{π}{2} - \frac{θ}{2}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?