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So I have tried solving the following, seemingly simple, power series. The solution should be that the series diverges, but I do not see how I would go about testing for convergence/divergence apart from a root test, which wouldn't help either by the looks of things (is this a false assumption?).

$$ \sum_{n=1}^{\infty} \left (\frac{1+2i}{\sqrt{5}} \right )^{n } $$

With ratio test:

$$ \lim_{n\rightarrow \infty} \left |\left (\frac{1+2i}{\sqrt{5}} \right )^{n+1}\left (\frac{\sqrt{5}}{1+2i} \right )^{n} \right |=\lim_{n\rightarrow \infty} \left | \frac{1+2i}{\sqrt{5}} \right | = \frac{1}{\sqrt{5}}\left | 1+2i \right |= \frac {\sqrt{5}}{\sqrt{5}}=1 $$

Which is inconclusive. From here on I do not see any alternative apart from the root test, which looks like it shouldn't help.

Best regards, Raoul

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What is the absolute value of n'th term? Check it. You might be surprised.

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  • $\begingroup$ You might just as well start writing \to instead of \rightarrow. Also, why $5^n$ in the numerator? Is $|1+2i|=5$? Also, I didn't say a word about limit; just the n'th term. $\endgroup$ – Ivan Neretin Sep 28 '15 at 18:05
  • $\begingroup$ Thank you for the tip Ivan. I've noticed that I've made some unnecessary mistakes while concentrating on LaTeX.. $\endgroup$ – Raoul Sep 28 '15 at 18:06
  • $\begingroup$ To look at the n'th term, do I not need to use limits to approximate it? $\endgroup$ – Raoul Sep 28 '15 at 18:09
  • $\begingroup$ No; why? Why approximate when you have the exact formula for it: $a_n=\left(1+2i\over\sqrt5\right)^n$. Now, what is the absolute value? $\endgroup$ – Ivan Neretin Sep 28 '15 at 18:11
  • $\begingroup$ Ah yes. Do you mean as follows: $$ \left | a_{n} \right |= \frac {\left | \left( 1+2i \right)^{n} \right |}{\sqrt{5}^{n}} $$ And since $ \left( 1+2i \right)^{n} \to \infty, \left| a_{n}\right| \to \infty $ I apologise, but I find series in general quite difficult. $\endgroup$ – Raoul Sep 28 '15 at 18:18

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