Divergent Series/Examples/n over n^2 + i
Jump to navigation
Jump to search
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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(ii)}$