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I want to show that $$\sum\limits_{n=1}^{\infty}\frac{i^n}{\sqrt{n}}$$ is convergent, but not absolutely convergent.

Demonstrating that it is not absolutely convergent is easy since $$\left|\frac{i^n}{\sqrt{n}} \right|=\frac{1}{\sqrt{n}}$$ but $$\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n}}$$ diverges. I'm stuck showing that it is conditionally convergent.

5xum
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emka
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    chop the series in the series into the pairly and the impairly idexed addends. then use twice http://en.wikipedia.org/wiki/Alternating_series_test. for the non absolute convergence just consider the pairly indexed addends. – Max Mar 18 '14 at 09:04

2 Answers2

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We have

$$\left|\sum_{k=1}^n i^k\right|=\left|\frac{1-i^n}{1-i}\right|\le\frac{1+|i^n|}{\sqrt 2}=\sqrt2$$ and the sequence $\left(\frac1{\sqrt n}\right)_n$ is decreasing to $0$ so by the Dirichlet's test the given series is convergent and since it's not absolutely convergent as you proved it then this series is conditionally convergent.

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

But the decreasing sequences $\{\frac{1}{\sqrt{2n}}\}_n$ and $\{\frac{1}{\sqrt{2n-1}}\}_n$ converge to $0$ as $n \rightarrow \infty$. Hence by alternating series test, we get that right hand side of the above equation converges.

Carl Butcher
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