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


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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