# Convergence/Divergence of Complex Series $\sum\limits_{n=1}^{\infty} \frac{n(2+i)^n}{2^n}$

$$\sum\limits_{n=1}^{\infty} \frac{n(2+i)^n}{2^n}$$

My Attempt: I am new to analyzing complex series, so please forgive me in advance. I apply the ratio test:

$$\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty}\frac{|(n+1)(2+i)^{n+1}2^n|}{|2^{n+1}n \ (2+i)^n|} = \lim_{n \to \infty} |\frac{n+1}{2n}(2+i)| = \frac{1}{2} \lim_{n \to \infty} |2+i|$$

I know that $$|z| = |a + bi|$$ can be expressed as $$\sqrt{a^2+b^2}$$, hence:

$$\frac{1}{2} \lim_{n \to \infty} \sqrt{5} > 1$$ By the ratio test, this makes the series diverging series. Is this approach correct?

• Your series and your ratio test don't match up. Aug 2 '20 at 23:09
• My apologies, an edit has been made: $(2+i)^n$ Aug 2 '20 at 23:17
• The final conclusion is wrong. The series is divergent. Aug 2 '20 at 23:19

Your approach is good. You can alternatively solve it through the root test. One has that \begin{align*} |a_{n}| = \left|\frac{n(2+i)^{n}}{2^{n}}\right| = n\left(\frac{\sqrt{5}}{2}\right)^{n} \Longrightarrow \limsup_{n\to\infty}|a_{n}|^{1/n} = \limsup_{n\to\infty}n^{1/n}\left(\frac{\sqrt{5}}{2}\right) = \frac{\sqrt{5}}{2} > 1 \end{align*}

Thus the given series diverges.

Hopefully this helps.

The ratio test claims that when $$\lim_{n\to\infty} |\frac{a_{n+1}}{a_{n}}|<1$$ the series $$\sum_{n=1}^{\infty} a_{n}$$ converges absolutely and when $$\lim_{n\to\infty} |\frac{a_{n+1}}{a_{n}}|>1$$ the series $$\sum_{n=1}^{\infty} a_{n}$$ diverges. As $$\frac{\sqrt{5}}{2}>1$$ the series diverges.

• $\left| \frac{a_{n+1}}{a_n} \right| = \frac{n+1}{n} \left| \frac{2+i}{2} \right| >1.$
– mjw
Aug 2 '20 at 23:25

Note that $$|z|\ge |\Re(z)|$$, and so $$\left|n \left(\frac{2+i}{2}\right)^n\right|=n \frac{|2+i|^n}{2^n}\ge n \frac{|\Re(2+i)|^n}{2^n}=n,$$so the series fails the test for divergence.