# Determining whether $\sum_{n = 0}^{\infty} 4\cos(2\pi n)e^{-3n}$ diverges

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!

• Did you tried to see if it was absolutely convergent? Or Root test? – Aram Nov 4 '14 at 5:25
• I haven't learnt about the root test ): – Joshua Nov 4 '14 at 5:29
• It's okay Timbuc posted a far simpler answer – Aram Nov 4 '14 at 5:30
• Converges: The factor ${\rm e}^{-3n}$ makes the job. – Felix Marin Nov 10 '14 at 21:37

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 $\dfrac{4}{1-e^{-3}}$.
$$\left|4\cos2\pi n\;e^{-3n}\right|\le\frac4{e^{3n}}=4\left(\frac1{e^3}\right)^n$$