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Could you tell me how to determine convergence of series with terms being products of real and complex numbers, like this:

$\sum_{n=1} ^{\infty}\frac{n (2+i)^n}{2^n}$ , $ \ \ \ \ \sum_{n=1} ^{\infty}\frac{1}{\sqrt{n} +i}$?

I know that $\sum (a_n +ib_n)$ is convergent iff $\sum a_n$ converges and $\sum b_n$ converges.

(How) can I use it here?

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Well, if you can prove absolute convergence, for instance... –  J. M. Apr 13 '13 at 13:49

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up vote 2 down vote accepted

Just like for series of reals (the underlying theorems are in the complex numbers, after all).

Your first series doesn't converge. By the ratio test, as $n \rightarrow \infty$: $$ \lvert \frac{(n + 1) (2 + i)^{n + 1}/2^{n + 1}}{n (2 + i)^n / 2^n} \rvert = \lvert \frac{(n + 1) (2 + i)}{2 n} \rvert = \frac{(n + 1) \sqrt{5}}{2 n} \rightarrow \frac{\sqrt{5}}{2} $$

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We have $$\left|\frac{n(2+i)^n}{2^n}\right|=n\left(\frac{\sqrt{5}}{2}\right)^n\not\to0$$ so the series $\displaystyle\sum_{n=1}^\infty \frac{n(2+i)^n}{2^n}$ is divergent.

For the second series we have $$\frac{1}{\sqrt{n}+i}\sim_\infty\frac{1}{\sqrt{n}}$$ then the series $\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}+i}$ is also divergent.

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Thank you. Could you tell me if I can use Cauchy's root test for $(\frac{n(2-i)+1}{n(3-2i)-3i})^n$? –  Andrew Apr 13 '13 at 14:36
    
@Andrew Yes, if we denote by $u_n$ your given expression then we have $$\sqrt[n]{|u_n|}=\left|\frac{n(2-i)+1}{n(3-2i)-3i}\right|=\sqrt{\frac{5n^2+4n+1‌​}{13n^2+12n+9}}\to\sqrt{\frac{5}{13}}<1$$so the series is convergent. –  Sami Ben Romdhane Apr 13 '13 at 15:24

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