Series are one of the most frustrating topics in calculus. They can be very confusing, and it can be very hard to determine if a series converges or not. If that weren’t enough, there are two types of convergence – absolute and conditional. This article will take you through what conditional convergence is and how to determine if a series is conditionally convergent.

### The Alternating Series Test

In order for us to learn about conditional convergence, we need to learn about the Alternating Series Test as most conditionally convergent series are alternating. The alternating series test states that for any alternating series $\displaystyle\sum_{n=0}^{\infty} a_n$ converges if there exists some $K$ such that for all $n > K$, $\left\|a_n\right\| < \left\|a_{n+1}\right\|$, and also that $\displaystyle\lim_{n\to\infty} \left\|a_n\right\| = 0$. In other words, an alternating series is convergent if there is some point where the absolute value of each term in the series decreases as $n$ gets greater and the terms in the series get closer and closer to 0. This test does not work on non-alternating series, though. For example, the Harmonic Series $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n}$ does not converge even though the absolute value of each term decreases as $n$ gets greater and the terms in the series get closer and closer to 0. A fuller description of the alternating series test can be found here {insert link}.

### What is Conditional Convergence?

Conditional convergence is a type of convergence in which the sum of a series converges, but the sum of the absolute values of the terms does not. A classic example is the series $\displaystyle\sum_{n=1}^\infty (-1)^{n+1}\cdot\dfrac{1}{n}$. You can see that this fulfills all of the conditions of the alternating series test, but the absolute value of that series, $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n}$, diverges. Be careful and make sure that both the alternating series converges and the absolute value diverges, though. For example, $\displaystyle\sum_{n=1}^\infty (-1)^n$ is not conditionally convergent, even though the absolute value of the series does diverge, because the original series does not converge. Similarly, $\displaystyle\sum_{n=1}^\infty (-1)^{n+1}\cdot\dfrac{1}{n^2}$ is not conditionally convergent because, although the alternating series converges, the sum of the absolute values of the terms, $\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}$, converges as well.

### Why Does This Matter?

The reason absolute and conditional convergence matter is because conditionally convergent series can actually be rearranged to get different sums! For example, let us write $\displaystyle\sum_{n=1}^\infty (-1)^{n+1}\cdot\dfrac1n$ as $1 – \dfrac12 + \dfrac13 – \dfrac14 + \dfrac15 – \dfrac16+ …$. We already proved that this series is conditionally convergent, and it turns out that this specific ordering converges to $\ln(2)$, although we won’t provide a proof here. However, we can rearrange the terms as $(1 – \dfrac12) – \dfrac14 + (\dfrac13 – \dfrac16) – \dfrac18 + (\dfrac15 – \dfrac{1}{10}) – \dfrac{1}{12}+…$, which turns out to be $\dfrac12 – \dfrac14 + \dfrac16 – \dfrac18 + \dfrac{1}{10} – \dfrac{1}{12}+…$, which is only half of our original sum! We didn’t leave any terms out; every odd $n$ appears in parentheses with $2n$, while every $n$ divisible by 4 is outside the parentheses. And it turns out that we can rearrange any conditionally convergent series to be any number! This makes more sense if we consider the series as a sum of two series, one with all positive numbers and one with all negative numbers. If we want to make the series sum to some constant $k$, all we do is sum the terms of the positive series (which, because the series is conditionally convergent, diverges to infinity) until we get a sum greater than $k$, then add the terms of the negative series (which diverges to negative infinity) until the sum becomes less than $k$, and then add the numbers of the positive series until the sum is greater than $k$ again, and so on. Because the absolute values of the terms of the series get smaller and smaller, after each such step the amount by which the sum differs from $k$ will also get smaller and smaller, and thus our series will converge to $k$. This whole process can’t be done with absolutely convergent series, because the sums of the positive and negative series won’t diverge to infinity. For example, for the series $\displaystyle\sum_{n=1}^\infty (-1)^{n}\cdot \dfrac{1}{n^2}$ (which is absolutely convergent), the positive series is $\displaystyle\sum_{n=1}^\infty \dfrac{1}{(2n)^2}$, which is less than the series $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ (try writing the terms out if you aren’t convinced). By the $p$-series test, $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, so the positive part of $\displaystyle\sum_{n=1}^\infty (-1)^{n} \dfrac{1}{n^2}$ also converges (the negative part of the series also converges by the same logic), and thus we cannot rearrange the terms to get whatever number we want, because no matter what the answer will just be the finite sum of the positive series plus the finite sum of the negative series.

### Conclusion

You should now know a lot more about alternating series, the tests that are needed to tell if they converge or diverge, and the difference between absolute and conditional convergence. Good luck on your future math journeys!