Question

Let \(\sum b_{n}\) be obtained by rearranging finitely many terms of a convergent series \(\sum a_{n} .\) Show that the two series have the same sum.

Short answer

In summary, given a convergent series \(\sum a_n\) and its rearranged series \(\sum b_n\), we proved that both series have the same sum, even after rearranging finitely many terms. This was done by analyzing the partial sums of the two series, showing that the sum of the terms beyond a certain index is the same for both series, and relating the sums of the first few terms using rearranged indices. This proof demonstrates the unique properties of convergent series and how a limited rearrangement of terms does not affect the final sum of the series.

Step-by-step solution

Define the given series and their partial sums

Let \(\sum a_{n}\) be a convergent series with partial sums \(s_n\). Similarly, let \(\sum b_{n}\) be the rearranged series of \(\sum a_{n}\) with partial sums \(t_n\). Remember that by the definition of convergence, \(\lim_{n\to\infty} s_n = s\) for some finite value \(s\).

Rearrange finitely many terms

Since we are only allowed to rearrange finitely many terms, let's assume that we rearrange terms up to index \(k\) such that \(a_1, a_2, ..., a_k\) are rearranged to \(b_1, b_2, ..., b_k\). The series after index \(k\) remains the same. Thus, we have: \(a_{k+1} = b_{k+1}, a_{k+2} = b_{k+2}, \dots\).

Compare partial sums

First, note that since we have only rearranged the first \(k\) terms, the partial sums of \(\sum a_n\) and \(\sum b_n\) beyond index \(k\) will be the same. That is, \(s_{k+n} - s_k = t_{k+n} - t_k\) for any \(n \in \mathbb{N}\). In other words, both series have the same sum from index \(k+1\) to \(\infty\). Next, observe that the partial sums of the first \(k\) terms of the rearranged series involve the same numbers but in a different order: \(t_k = b_1+b_2+\dots+b_k = a_{\pi(1)}+a_{\pi(2)}+\dots+a_{\pi(k)}\). We also know that adding the first \(k\) terms of any series corresponds to the difference between the \(k\)-th and \((k-1)\)-th partial sums; i.e., \(s_{k} - s_{k-1} = a_1 + a_2 + \dots + a_k\). Therefore, \(s_{k}-s_{k-1}=t_k-t_{k-1} \Rightarrow s_{k}-s_{k-1}= a_{\pi(1)}+a_{\pi(2)}+\dots+a_{\pi(k)}\).

Prove the two series have the same sum

Finally, we want to show that the limit of the partial sums of both series is the same, which would prove that the two series have the same sum. Using the fact that both series have the same sum from index \(k+1\) to \(\infty\) and the relationship between the first \(k\) terms, we have: \(s = \lim_{n\to\infty} s_n = \lim_{n\to\infty}(s_k + (s_{k+n} - s_k)) = s_k + \lim_{n\to\infty}(t_{k+n} - t_k) = t_k + \lim_{n\to\infty}(t_{k+n} - t_k)\). Since \(t_k = s_{k}-s_{k-1} = a_{\pi(1)}+a_{\pi(2)}+\dots+a_{\pi(k)}\), our equation becomes: \(s = a_{\pi(1)}+a_{\pi(2)}+\dots+a_{\pi(k)} + \lim_{n\to\infty}(t_{k+n} - t_k) = \lim_{n\to\infty}(t_k + (t_{k+n} - t_k)) = \lim_{n\to\infty} t_n\). Thus, we have proved that \(\sum a_n\) and \(\sum b_n\) have the same sum, which we denoted as \(s\).

Key concepts

Partial Sums

When working with series, a helpful concept to understand is **partial sums**. A partial sum, denoted as \(s_n\), is simply the sum of the first \(n\) terms of a series. For a series \(\sum a_{n}\), its partial sum \(s_n\) would be expressed as:

• \(s_1 = a_1\)
• \(s_2 = a_1 + a_2\)
• \(s_3 = a_1 + a_2 + a_3\)
• ... and so on.

The idea is to see how the series builds up as we add more terms. If a series converges, the partial sums will approach a specific number, known as the sum of the series. To visualize convergence, think of walking towards a line; as you take more steps (add more terms), you get closer to this line (the sum). This concept is crucial in proving equal sums after series rearrangement.

Rearrangement of Series

Rearranging a series is akin to reordering its terms. For some series, reordering might change the sum, especially in cases of conditionally convergent series. However, in this exercise, we only rearrange finitely many terms. This means after a certain position, the terms in both the original and new series align, i.e., \(a_{k+1} = b_{k+1}, a_{k+2} = b_{k+2}, \ldots\). This limited rearrangement ensures that changes to the series do not modify the overall sum.
In essence, think of a jigsaw puzzle; if you rearrange a few pieces but keep the remaining part intact, the overall picture remains unchanged. This stability allows the sum of a convergent series and its rearranged counterpart to remain equal.

Sum of Series

The **sum of a series** refers to the value that the partial sums \(s_n\) approach as \(n\) becomes very large. For a convergent series \(\sum a_{n}\), there is a specific number, \(s\), such that \(\lim_{n\to\infty} s_n = s\).
In simpler words, if you keep adding terms forever, you'll approach this number \(s\) but never actually surpass it. The intriguing part about convergent series is that even if minor rearrangements are made initially, the series still approaches the same sum. This is because the essence of convergence relies on the infinite continuation beyond those first few terms, thereby retaining the stability of \(s\).

Limits of Sequences

The concept of 'limit' is central to understanding series and their convergence. In mathematical terms, the **limit of a sequence** is the value that the terms of a sequence get closer to as the sequence progresses. This essentially tells us about long-term behavior.
Consider a sequence formed by the partial sums of a series. As you add more and more terms to the series, you form a sequence like \([s_1, s_2, s_3, \dots]\). The limit of this sequence of partial sums is what we refer to as the sum of the series.

• If \(\lim_{n\to\infty} s_n = s\), then the series is convergent, and \(s\) is its sum.
• If the sequence doesn't settle towards a single value, the series is divergent.

This property of limits ensures that despite rearranging finitely many terms of a convergent series, the fundamental behavior and primary destination—its limit—remain unaffected, and the sum stays the same.