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I'd like to know whether the complex series

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

is convergent or not.

I guess it is divergent, because in order for complex series $c_n=a_n+i(b_n)$ to be convergent, sum of $a_n$ and sum of $b_n$ should both converge, but sum of $a_n$ = sum of $\frac{\log n}{n}$ is divergent. So it's divergent.

Is my argument right? if there's some wrong point, could you point it out?

thank you for your comment in advance and sorry for bad mathematical writing, since I wrote this on my phone.

Dando18
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glimpser
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  • Did you mean $\sum_{n=1}^\infty \left( \frac{\log n}{n} + i^n(\frac{\log n}{n})\right)$ or something else??? – Dando18 Jul 22 '17 at 05:01

1 Answers1

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The argument as stated is wrong because $a_{n}$ and $b_{n}$, the real and imaginary parts of the series, are not $\frac{\log{n}}{n}$. In fact, consider the 2nd term, $c_{2} = \frac{\log{n}}{n} + i^{2}\frac{\log{n}}{n} = 0$. Depending on the value of $n\mod{4}$, $c_{n}$ is $c_{1} = \frac{\log{n}}{n} + i\frac{\log{n}}{n}$, $c_{2} = 0$, $c_{3} = \frac{\log{n}}{n} - i\frac{\log{n}}{n}$ or $c_{4} = \frac{2\log{n}}{n}$.

Here is another idea: it is sufficient to show that the real part does not converge to show that the entire series does not. From the above, I suspect that you can show that $a_{n} \geq \frac{\log{n}}{n}$, meaning that $a_{n}$ does not converge and so $c_{n}$ does not. This detail is left to you (thought it could be false, I'm just eyeballing.)

Wraith1995
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