Divergent Series/Examples/n over n^2 + i

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Example of Divergent Series

The complex series defined as:

$\ds S = \sum_{n \mathop = 1}^\infty \dfrac n {n^2 + i}$

is divergent.


Proof

\(\ds S\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac n {n^2 + i}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {n \paren {n^2 - i} } {\paren {n^2 + i} \paren {n^2 - i} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {n^3 - i n^2} {n^4 + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac {n^3} {n^4 + 1} - \frac {i n^2} {n^4 + 1} }\)


Then we have:

\(\ds \sum_{n \mathop = 1}^\infty \frac {n^3} {n^4 + 1}\) \(\ge\) \(\ds \sum_{n \mathop = 1}^\infty \frac {n^3} {2 n^4}\)
\(\ds \) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 1}^\infty \frac 1 n\)


From Harmonic Series is Divergent, $\ds \sum_{n \mathop = 1}^\infty \dfrac 1 n$ is a divergent series.

The result follows from Convergence of Series of Complex Numbers by Real and Imaginary Part.

$\blacksquare$


Sources