Infinite series with cos in numerator

Question

How do you evaluate this series?

$$\sum_{i=1}^{\infty}\frac{\cos i}{2^i}$$

It's absolutely convergent by comparison to the geometric series. But the $\cos$ is tripping me up. I've tried differentiating in order to go through the $\cos$ -> $\sin$ -> $\cos$ route, but that gives me different powers of 2 in the denominator. Any ideas?

Write $\cos x$ as $\frac{e^{ix} + e^{-ix}}{2}$ (your use of $i$ as the indexing variable is very unfortunate for this reason). Also, what are you differentiating? — Qiaochu Yuan, Jul 28 '16 at 19:59
Thanks. I was differentiating w.r.t. i inside the sum, but in retrospect that's nonsense. — Number 34, Jul 28 '16 at 21:17

score 5 · Answer 1 · answered Jul 28 '16 at 20:00

5

Hint. One may write, for $|q|<1$, $\alpha \in \mathbb{R}$, $$ \sum_{n=1}^{\infty} q^n \cos(n\alpha)=\Re \sum_{n=1}^{\infty} (qe^{i\alpha})^n =\Re\: \frac{qe^{i\alpha}}{1-qe^{i\alpha}} $$ where we have used the standard evaluation of a geometric series.

answered Jul 28 '16 at 20:00

Olivier Oloa

120,989

1

Darn, beat me to it. (As a side remark: here was used the fact that the real-part is linear, i.e. $\mathfrak{R}\sum_{k=1}^\infty a_k = \sum_{k=1}^\infty \mathfrak{R}a_k$.) – Clement C. Jul 28 '16 at 20:02
That makes sense, thanks. So easy once you see how. – Number 34 Jul 28 '16 at 21:18
@Number 34 You are welcome. – Olivier Oloa Jul 28 '16 at 21:20

score 5 · Accepted Answer · answered Jul 28 '16 at 20:07

5

$$ S=\sum_{n\geq 1}\frac{\cos n}{2^n}=\text{Re}\sum_{n\geq 1}\left(\frac{e^i}{2}\right)^n = \text{Re}\left(\frac{e^i}{2-e^i}\right)=\text{Re}\left(\frac{2e^i-1}{5-4\cos 1}\right)=\color{red}{\frac{2\cos 1-1}{5-4\cos 1}}. $$

answered Jul 28 '16 at 20:07

Jack D'Aurizio

353,855

That's great, thanks. You've made it very clear. – Number 34 Jul 28 '16 at 21:20

Infinite series with cos in numerator

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