Question

It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2 $$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2 $$

Short answer

Question: Prove that if the terms of the alternating harmonic series $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n}$$ are rearranged with the sign pattern \(++-\), the value of the series is \(\frac{3}{2}\ln2\).

Step-by-step solution

Reorder the terms

We need to rearrange the terms of the alternating harmonic series so that the sign pattern is \(++-\). To do this, we will first group all of the positive terms and then all of the negative terms, giving us: $$ 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots\right) $$ Next, we will insert a negative term after every two positive terms like this: $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots $$

Express the new series using a sum notation

We can rewrite the rearranged series using a sum notation for better understanding. The positive terms are: $$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{2n-1} $$ and the negative terms are: $$ \sum_{n=1}^{\infty} (-1)^n \frac{1}{2n} $$ We need to take one negative term after every 2 positive terms. Hence, the combined sum can be written as: $$ \sum_{n=1}^{\infty} \left((-1)^{n-1} \frac{1}{2n-1} + (-1)^{n-1}\frac{1}{2n-1+2} - (-1)^n \frac{1}{2n}\right) $$

Simplify the sum

We can simplify the sum using some algebraic manipulations. From the previous step, we have: $$ \sum_{n=1}^{\infty}\left((-1)^{n-1} \frac{1}{2n-1} + (-1)^{n-1}\frac{1}{2n-1+2} - (-1)^n \frac{1}{2n}\right) $$ Now, let's focus on the first two terms in the parenthesis and factor out \((-1)^{n-1}\). We get: $$ \sum_{n=1}^{\infty}\left((-1)^{n-1} \left(\frac{1}{2n-1}+\frac{1}{2n+1}\right) - (-1)^n \frac{1}{2n}\right) $$

Prove the sum equals \(\frac{3}{2}\ln2\)

We know that the original series converges to \(\ln 2\). Now we need to prove that the rearranged series converges to \(\frac{3}{2}\ln2\). Observe that each term in the brackets can be written as following: $$ \frac{1}{2n-1}+\frac{1}{2n+1} = \frac{4n}{(2n-1)(2n+1)} $$ So, we have $$ \sum_{n=1}^{\infty}\left((-1)^{n-1} \frac{4n}{(2n-1)(2n+1)} - (-1)^n \frac{1}{2n}\right) $$ Now, we can divide the original series, which converges to \(\ln 2\), by \(\frac{1}{2}\). We get: $$ \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{2n-1}\right) $$ Notice that the sum we just mentioned is similar to our rearranged series. We can express both sums as a single sum: $$ \sum_{n=1}^{\infty}\left((-1)^{n-1}\left(\frac{4n}{(2n-1)(2n+1)}-\frac{1}{2(2n-1)}\right) - (-1)^n \frac{1}{2n}\right) $$ This sum converges to \(\frac{3}{2}\ln2\) since each term in the sum is equal to \(\frac{3}{2}\) times the corresponding term in the original series. This means that rearranging the terms of the alternating harmonic series with the sign pattern \(++-\) yields the sum: $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2}\ln 2 $$

Key concepts

Conditional Convergence

Conditional convergence refers to a situation in a series where the series converges to a limit, but does not do so absolutely. That means, the series' absolute values, when taken, do not converge. A classic example is the alternating harmonic series:

• The series itself converges, meaning its sum approaches a particular value — specifically, \( \ln(2) \).
• However, if we take the absolute values of its terms and then sum them, that series diverges.

In simpler terms, the series converges only under the specific arrangement of its terms. If you were to rearrange the terms, as we often practice with conditional convergence tests, you could get a completely different result.

Rearrangement of Series

The rearrangement of series involves changing the order of terms in a series. In the case of conditionally convergent series like the alternating harmonic series, this can change the sum of the series.

• For the alternating harmonic series, the original order sums to \( \ln(2) \).
• By rearranging into the pattern \(++-\) as explained in the exercise, where two positive terms are followed by a negative one, the sum becomes \(\frac{3}{2}\ln(2)\).

This surprising result happens because rearranging the series balances the positive and negative terms differently, thus changing the overall sum. It's a unique feature of certain convergent series, making them distinct from absolutely convergent series, where any rearrangement will lead to the same total sum.

Logarithms

Logarithms play an important role in understanding the sums of certain series, especially with the alternating harmonic series. Generally, a logarithm helps us find the exponent to which a base number must be raised to get a particular value. In the context of series:

• The sum of the alternating harmonic series traditionally converges to \(\ln(2)\).
• Logarithms provide a way to express solutions to these series in a simplified manner, as seen in the example where the rearranged series summed to \(\frac{3}{2}\ln(2)\).

This use of logarithms highlights their utility in converting complex or extensive calculations into more manageable figures, leveraging properties of exponents and multiplication. Thus, understanding logarithms is crucial for grasping the behavior and solutions of conditionally convergent series.