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Consider the following infinite series: $$ \displaystyle \sum_{n = 0}^{\infty} 4\cos(2\pi n)e^{-3n} $$ Determine whether the infinite series diverges or converges.

I tried to use:

  1. The Integral Test -- failed few conditions, since I realized that $f$ is not always decreasing (it oscillates)
  2. The Divergence Test -- also not helpful, since $\displaystyle \lim_{n \to \infty} 4\cos(2\pi n)e^{-3n} = 0$

Any suggestions or hints? Thanks!

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  • $\begingroup$ Did you tried to see if it was absolutely convergent? Or Root test? $\endgroup$ – Aram Nov 4 '14 at 5:25
  • $\begingroup$ I haven't learnt about the root test ): $\endgroup$ – Joshua Nov 4 '14 at 5:29
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    $\begingroup$ It's okay Timbuc posted a far simpler answer $\endgroup$ – Aram Nov 4 '14 at 5:30
  • $\begingroup$ Converges: The factor ${\rm e}^{-3n}$ makes the job. $\endgroup$ – Felix Marin Nov 10 '14 at 21:37
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Note that $\cos(2\pi n) = 1$ for every integer $n$. Thus, $\displaystyle\sum_{n=0}^{\infty}4\cos(2\pi n)e^{-3n} = \sum_{n=0}^{\infty}4e^{-3n}$ which is a geometric series whose common ratio is $e^{-3} \in (0,1)$. Thus, the sum converges to $\dfrac{4}{1-e^{-3}}$.

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$$\left|4\cos2\pi n\;e^{-3n}\right|\le\frac4{e^{3n}}=4\left(\frac1{e^3}\right)^n$$

and by the comparison test not only does the series converges but it also converges absolutely.

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