# Does the attached series converge or diverge? Note: it is a complex series and not a real series. I've been told the limit comparison test is not a valid approach to solving this problem. sum_{n=1}^inftyfrac{1}{n-i^n}

Does the attached series converge or diverge? Note: it is a complex series and not a real series. Ive
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Laaibah Pitt

The given series is
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n-{i}^{n}}$
Let us solve the given problem by the help of the Ratio test.
The Ratio test is,
For a series
Then, the series is convergent if $L<1$
the series is divergent if $L>1$
And if $L=1$, we have no result or we can say the test fails.
Here ${a}_{n}=\frac{1}{n-{i}^{n}}$
So, $\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|=\underset{n\to \mathrm{\infty }}{lim}|\frac{n-{i}^{n}}{\left(n+1\right)-{i}^{n+1}}|\left(\frac{\mathrm{\infty }}{\mathrm{\infty }}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{1-{i}^{n}\mathrm{ln}i}{1-{i}^{n+1}\mathrm{ln}i}|\left(\frac{\mathrm{\infty }}{\mathrm{\infty }}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{{i}^{n}\left(\mathrm{ln}i{\right)}^{2}}{-{i}^{n+1}\left(\mathrm{ln}i{\right)}^{2}}|$
$=\frac{1}{i}$
$=\frac{1}{i}\left(\frac{i}{i}\right)$
=-i,(as ${i}^{2}=-1$)
Therefore, $L=-i<1$
So, the given series is convergent.
ANSWER: The given series is convergent.

Jeffrey Jordon