Compute $\sum\limits_{n=1}^{\infty}\frac{\cos n}{2^n}$

Question

I have some problems when calculating the sum of this infinite series: $$ \sum _{n=1}^{\infty}\frac{\cos n}{2^n} $$ I've checked the convergence and found that $\sum _{n=1}^{\infty}\frac{\cos n}{2^n}$ converges (absolutely).

The answer is: $\sum _{n=1}^{\infty}\frac{\cos n}{2^n}= \frac{2cos1−1}{5−4cos1}$.

Related : https://math.stackexchange.com/questions/2601940/is-there-an-easy-way-to-calculate-sum-n-020-frac-sinn-pi-402n — lab bhattacharjee, Feb 01 '18 at 06:49

score 6 · Accepted Answer · answered Jan 31 '18 at 22:07

6

Your sum is the real part of the geometric series $$\sum_{n=1}^\infty\frac{e^{in}}{2^n}=\sum_{n=1}^\infty\left(\frac{e^i}2\right)^n=\frac{e^i}{2-e^i}=\frac{e^i(2-e^{-i})}{(2-e^i)(2-e^{-i})} =\frac{2e^i-1}{5-4\cos 1}.$$

answered Jan 31 '18 at 22:07

Angina Seng

158,341

Compute $\sum\limits_{n=1}^{\infty}\frac{\cos n}{2^n}$

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