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Analyze the convergence of the series $\sum_{n=1}^{\infty} \frac{\cos (in)}{2^{n}}$.

$a_n=\frac{1}{2^n}, \quad b_n=\cos(in), \quad \lim(a_{n})=0$ but $\lim(b_{n})$ does not exist. So I think the series diverges. How do I prove it formally?

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  • $\begingroup$ $$i=\sqrt{-1}$$?? $\endgroup$ – lab bhattacharjee Jun 16 '17 at 9:48
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    $\begingroup$ If $i$ is supposed to be the imaginary unit, observe $\cos ix=\cosh x.$ And correct the spelling, please, "analyze", there could be misunderstandings. $\endgroup$ – Professor Vector Jun 16 '17 at 9:53
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Observe that $$ \frac{\cos(in)}{2^n} = \frac{\cosh(n)}{2^{n}} = \frac{e^{-n} + e^{n}}{2^{n+1}}.$$ Now $\frac{e^{-n}}{2^{n+1}} \to 0$ and $\frac{e^{n}}{2^{n+1}}$ does not converge to 0 (you may check that $f(x) = \frac{e^x}{2^{x+1}}$ has positive derivative for every $x>0$) and hence $\frac{\cos(in)}{2^n}$ does not converge to 0 and the series diverges.

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Indeed,

$$\frac{e^{i^2n}+e^{-i^2n}}{2^{n+1}}$$ yields the sum of two geometric series of ratios $\dfrac1{2e}$ and $\dfrac e2$, and the second diverges.

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