Rearrangement of Terms in Convergent Series

# Rearrangement of Terms in Convergent Series

Thus far we have looked at a series that converges to have only one sum. We will now note that if a series is conditionally convergent, then rearranging the terms in this series will allow us to change the sum of that series.

 Theorem 1: If $\sum_{n=1}^{\infty} a_n$ is an absolutely convergent series to the sum $s$, then any rearrangement of the terms of this series will also converge to the sum $s$. If $\sum_{n=1}^{\infty} a_n$ is a conditionally convergent series to the sum $s$, then the terms of this series can be rearranged to converge to any $s \in \mathbb{R}$ or to diverge to $\pm \infty$.

The theorem above tells us that the sum to which a conditionally convergent series converges to is dependent on the order of the terms of the series. Let's now look at an example of rearranging terms in a conditionally convergent series.

First consider the alternating harmonic series, $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ which we've verified to already be a conditionally convergent sequence as it passes all of the requirements of the alternating series test. If order is preserved in this series, then the sum of this series will be $\ln 2$ which will be verified later. Suppose that instead we want to change the order of the terms so that this series converges to $2$ instead. We can do that by the theorem given above.

The first step if to sum up all of the positive terms of the series until we reach or surpass $2$:

(1)
\begin{align} \quad \left ( 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right ) \approx 2.02... \end{align}

Now we will add as many negative terms as necessary until we get below $2.02...$:

(2)
\begin{align} \quad \left ( 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right ) - \left ( \frac{1}{2} \right ) \approx 1.52... \end{align}

Then we will continue to add as many positive terms of the series until we reach or surpass $2$ again:

(3)
\begin{align} \quad \quad \left ( 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right ) + \left ( -\frac{1}{2} \right ) + \left ( \frac{1}{17} + \frac{1}{19} + ... + \frac{1}{41} \right ) \approx 2.004... \end{align}

Once again we will add as many negative terms as necessary until we get below $2.004...$:

(4)
\begin{align} \quad \quad \left ( 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right ) + \left ( -\frac{1}{2} \right ) + \left ( \frac{1}{17} + \frac{1}{19} + ... + \frac{1}{41} \right ) + \left ( -\frac{1}{4} \right) \approx 1.754... \end{align}

We will repeat this step indefinitely, and the sum will converge to $2$ as desired.