Problem 2 - A Course in Calculus and Real Analysis

Question

Consider a series \(\sum_{k} a_{k}\) and for each \(n=0,1,2, \ldots\), let \(b_{n, k}:=a_{n+k} .\) Show that the series \(\sum_{k} a_{k}\) is convergent if and only if for some \(n=0,2, \ldots\), the series \(\sum_{k} b_{n, k}\) is convergent. In this case, prove that the series \(\sum_{k} b_{n, k}\) is convergent for every \(n=0,1,2, \ldots\), and \(\sum_{k} b_{n, k}=\sum_{k} a_{k}-A_{n}\), where \(A_{n}\) is the \(n\) th partial sum of the series \(\sum_{k} a_{k}\).

Step-by-Step Solution

Answer

In summary, we showed that the series \(\sum_{k} a_{k}\) is convergent if and only if for some \(n = 0, 1, 2, \ldots\), the series \(\sum_{k} b_{n, k}\) is convergent. Moreover, we proved that in this case, the following equality holds: \[\sum_{k} b_{n, k} = \sum_{k} a_{k} - A_n\]

Step 1: Define related series

First, we will define the related series, where \(b_{n, k} = a_{n+k}\) for \(n = 0,1,2,\ldots\).

Step 2: If: Show that if the series \(\sum_{k} a_{k}\) is convergent, then \(\sum_{k} b_{n, k}\) is convergent for some \(n = 0,2,\ldots\)

Assume that the series \(\sum_{k} a_{k}\) is convergent, which means there exists a limit \(L\) such that as \(n \to \infty\), the partial sum \(A_n \to L\). Now, consider the series \(\sum_{k} b_{n, k}\). Note that when \(n = 0\), the series becomes \(\sum_{k} a_{k}\), so it is convergent for at least one \(n \in \{0,2,3,\ldots\}\).

Step 3: Only if: Show that if \(\sum_{k} b_{n, k}\) is convergent for some \(n=0,2,\ldots\), then the series \(\sum_{k} a_{k}\) is convergent

Now, let's assume that the series \(\sum_{k} b_{n, k}\) is convergent for some \(n \in \{0,2,3, \ldots\}\). This means that it converges to a limit, let's call it \(L'\). We aim to show that the series \(\sum_{k} a_{k}\) is convergent. We know that \[ b_{n, k} = a_{n + k} \] Sum the series \(\sum_{k} b_{n, k}\) for \(k = 0, 1, 2, \ldots\). We can write it as \[\sum_{k} a_{n+k} = L'\] Split the series \(\sum_{k} a_{k}\) into two parts: one for \(k=0,1,2, \ldots, n-1\) and another for \(k=n, n+1, n+2, \ldots\). Then we have \[\sum_{k=0}^{n-1} a_k + \sum_{k=n}^{\infty} a_k = A_n + \sum_{k=n}^{\infty} a_k = A_n + L'\] Notice that the second part of the sum is equal to \(\sum_{k} b_{n, k}\). Therefore, the series \(\sum_{k} a_{k}\) converges to \(A_n + L'\).

Step 4: Equality: Prove that \(\sum_{k} b_{n, k}=\sum_{k} a_{k} - A_{n}\)

Now, let's show that \(\sum_{k} b_{n, k} = \sum_{k} a_{k} - A_n\). Recall that \[b_{n, k} = a_{n+k}\] Sum the series \(\sum_{k} b_{n, k}\) for \(k = 0, 1, 2, \ldots\). We can write it as \[\sum_{k} a_{n+k}\] As already mentioned above, we can express the series \(\sum_{k} a_{k}\) as \[A_n + \sum_{k=n}^{\infty} a_k\] By comparison, we have: \[\sum_{k} b_{n, k} = \sum_{k=n}^{\infty} a_k = \sum_{k} a_{k} - A_n\] This proves the required equality.