The complex series is $$\sum_{k=1}^{\infty} (i^k-\frac{1}{k^2})$$ I know the answer is that the series diverges.

Is this because $$\lim_{k\to\infty} (i^k-\frac{1}{k^2})$$ Does not exist since $i^k$ repeats itself infinitely? So it diverges by the kth term test for divergence?

  • $\begingroup$ Yes, because $|i|=1$, so the corresponding geometric series won't converge. Or the Test for Divergence, as you stated. $\endgroup$ – Integrand Mar 26 at 2:26
  • $\begingroup$ @Integrand Is it true that the sum of a divergence series, with a convergent series always diverges, and could i say, it is the sum of a divergent geometric series, with a convergent p-series? $\endgroup$ – user736276 Mar 26 at 2:28
  • 1
    $\begingroup$ Yep. If your series converged, you could split it up into two convergent series. Alas, one diverges and one converges, so yours diverges. $\endgroup$ – Integrand Mar 26 at 2:30

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