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

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?

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

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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.

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

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