# Convergence tests for a complex series?

For $\sum_{i=1}^{\infty} z_i$, where $z_i \in \mathbb{C}$, how should the convergence tests be performed?

I read somewhere that the tests applied for convergence of complex series are same as that for real series, but I am wondering if this is so, does that mean that imaginary part of the complex numbers doesn't play any role?

• Certainly absolute convergence of a complex series amounts to that of a real series. – hardmath Oct 17 '14 at 17:59
• that means that for series $\sum_{i=1}^{\infty} z_i$ to be convergent it is both necessary and sufficient that $\sum_{i=1}^{\infty} |z_i|$ be convergent ? – kaka Oct 17 '14 at 18:03
• We don't really know of any necessary and sufficient conditions for a real series to be convergent, so we can't know them for complex series either. Edit: Except for 'casting' the series into an equivalent problem, cauchy's test, etc. – Bruce Zheng Oct 17 '14 at 18:10
• For a complex series $\sum_{i=0}^\infty z_i$ to converge it is necessary and sufficient for the real parts $\sum_{i=0}^\infty \mathfrak{Re}(z_i)$ and the imaginary parts $\sum_{i=0}^\infty \mathfrak{Im}(z_i)$ (both these are real series) to converge. – hardmath Oct 17 '14 at 20:26
• @hardmath thanks. This is what I was interested in knowing! – kaka Oct 17 '14 at 20:46

Let me give more detail for my Comments above.

There are at least two senses in which the convergence of a complex series can be reduced to a question of real series convergence.

In the first sense we say "convergence" simply to mean having a limit, i.e. given $\epsilon \gt 0$, there exists $n$ s.t. all partial sums after the $n$th partial sum are within $\epsilon$ distance of the final limit.

Then for a complex series $\sum_{i=0}^\infty z_i$ to converge, it is necessary and sufficient for the real parts $\sum_{i=0}^\infty \mathfrak{Re}(z_i)$ and the imaginary parts $\sum_{i=0}^\infty \mathfrak{Im}(z_i)$ (both these are real series) to converge.

There are then some specialized notions. We say a series is absolutely convergent iff the series obtained by taking absolute values term by term converges:

$$\sum_{i=0}^\infty |z_i|$$

Whether the terms $z_i$ are real or complex, taking absolute values gives a series with nonnegative real terms. Furthermore, for either real or complex series, absolute convergence implies convergence in the first sense.

As it happens, absolute convergence of a complex series is equivalent to the absolute convergence of both the real parts and imaginary parts series already mentioned.

Conditional convergence commonly is used to mean a series that converges but does not converge absolutely. The alternating harmonic series is a well-known example of this with real numbers as terms.

In the case of complex series, the discussion above implies that a conditionally convergent complex series has conditional convergence of either the real parts or the imaginary parts series (or both).

Thus the convergence of complex series does not introduce new phenomena after convergence of real series, with the minor caveat that one could contrive a complex series whose real parts are conditionally convergent but whose imaginary parts are absolutely convergent, or vice versa.

Absolute convergence of complex series implies convergence. The common series tests for real series actually establish absolute convergence, so the ratio test, for example, carries over. But some complex series converge conditionally, just like real series. So this is not a necessary condition.

Example: $i\sum_{n\geq 1}(-1)^n\frac{1}{n}$ converges conditionally, but not absolutely in $\mathbb{C}$ and all of its partial sums are non-real.

• (+1) nice. So what should be the way to check the conditional convergence, then ? Both the real and imaginary parts should converge? – kaka Oct 17 '14 at 18:14
• Do you have an example of a series to examine? That would help, as my experience with non-absolutely converging series is case by case. – J. David Taylor Oct 17 '14 at 18:16
• $\sum_n \frac{e^{jLog(n)}+e^{-jLog(n)}}{n}$ – kaka Oct 17 '14 at 18:21
• So your series is the same as $\sum_n\frac{2\cos(\log(n))}{n}$. My approach would start with looking at the distribution of $\cos(\log(n))$. – J. David Taylor Oct 17 '14 at 19:45

There are tests for complex series that show convergence but not absolute converge when the series is not real. Take for example Abel's test. http://en.wikipedia.org/wiki/Abel%27s_test#Abel.27s_test_in_complex_analysis . It says that if $\sum_n a_n z^n$ diverges when $|z| > 1$ and converges when $|z| < 1$ ($z$ complex), and if $\lim_n a_n = 0$ and $a_1 > a_2 > a_3 \ldots$ where $a_n$ is a real sequence then $\sum_n a_n z^n$ converges for all $z$ with $|z| = 1$ except possibly $z = 1$.