Prove the summation

**mathwizard** · June 18th 2008, 11:16 AM

$\sum_{n=1} ^\infty \; \frac{\theta}{2^n}\;\tan\; \frac{\theta}{2^n} \;=\; \boxed{1 - \theta \;\cot\;\theta}$

**NonCommAlg** · June 18th 2008, 06:21 PM

Originally Posted by mathwizard

$\sum_{n=1} ^\infty \; \frac{\theta}{2^n}\;\tan\; \frac{\theta}{2^n} \;=\; \boxed{1 - \theta \;\cot\;\theta}$

start with this identity: $\tan \alpha = \cot \alpha - 2 \cot(2\alpha).$ so: $\frac{\theta}{2^j} \tan \left(\frac{\theta}{2^j} \right)=\frac{\theta}{2^j} \cot \left(\frac{\theta}{2^j} \right) - \frac{\theta}{2^{j-1}} \cot \left(\frac{\theta}{2^{j-1}} \right).$ thus:

$I_n=\sum_{j=1}^n \frac{\theta}{2^j} \tan \left(\frac{\theta}{2^j} \right)=\sum_{j=1}^n \left[\frac{\theta}{2^j} \cot \left(\frac{\theta}{2^j} \right)-\frac{\theta}{2^{j-1}} \cot \left(\frac{\theta}{2^{j-1}} \right) \right] \ \ \ \ \ \ \text{(telescoping sum)}$

$=\frac{\theta}{2^n} \cot \left(\frac{\theta}{2^n} \right) - \theta \cot \theta.$ therefore: $\lim_{n\to\infty} I_n= \lim_{n\to\infty} \frac{\theta}{2^n} \cot \left(\frac{\theta}{2^n} \right)- \theta \cot \theta=1 - \theta \cot \theta. \ \ \ \square$

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