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

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Example of Divergent Series

The complex series defined as:

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

is divergent.


Proof

\(\displaystyle S\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac n {n^2 + i}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {n \paren {n^2 - i} } {\paren {n^2 + i} \paren {n^2 - i} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {n^3 - i n^2} {n^4 + 1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {\dfrac {n^3} {n^4 + 1} - \dfrac {i n^2} {n^4 + 1} }\) $\quad$ $\quad$


Then we have:

\(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {n^3} {n^4 + 1}\) \(\ge\) \(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {n^3} {\paren {n + 1}^4}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 2}^\infty \dfrac {\paren {n - 1}^3} {n^4}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 2}^\infty \dfrac {n^3 - 3 n^2 + 3 n - 1} {n^4}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 2}^\infty \paren {\dfrac {n^3} {n^4} - \dfrac {3 n^2 + 3 n - 1} {n^4} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 2}^\infty \dfrac 1 n - \sum_{n \mathop = 2}^\infty \dfrac {3 n^2 + 3 n - 1} {n^4}\) $\quad$ $\quad$


From Harmonic Series is Divergent, $\displaystyle \sum_{n \mathop = 2}^\infty \dfrac 1 n$ is a divergent series.

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

$\blacksquare$


Sources