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Let \(\sum a_{n}\) be a given series and let \(\sum b_{n}\) be the series in which the terms are the same and in the same order as in \(\sum a_{n}\) except that the terms for which \(a_{n}=0\) have been omitted. Show that \(\sum a_{n}\) converges to \(A\) if and only if \(\sum b_{n}\) converges to \(A\).

Short Answer

Expert verified
A series \(\sum a_{n}\) and \(\sum b_{n}\), where \(\sum b_{n}\) omits the terms where \(a_{n}=0\), will both converge to the same value (in this case, \(A\)) if any one of them does. The zero terms in \(\sum a_{n}\) have no material impact on the limit.

Step by step solution

01

Define the Series

Given two series, \(\sum a_{n}\) and \(\sum b_{n}\), where series \(\sum b_{n}\) is same as \(\sum a_{n}\) but the terms for which \(a_{n}=0\) have been omitted. We need to prove that both the series converge to the same number A.
02

Assume Convergence

Let's start by assuming that \(\sum a_{n}\) converges to A. This implies that for every number \(ε > 0\), there is an integer \(N1\) such that \(| a_{n} - A | < ε \) for all \( n > N1\).
03

Prove \(\sum b_{n}\) Converges

Since all the non-zero terms of series \(\sum a_{n}\) are in series \(\sum b_{n}\), the convergence of \(\sum a_{n}\) to A implicates the same convergence for \(\sum b_{n}\), because \( a_{n} = b_{n} \) amiss those terms for which \( a_{n} = 0\). Hence, \(\sum b_{n}\) converges to A.
04

Assume \(\sum b_{n}\) Converges

On the other hand, assume that \(\sum b_{n}\) converges to A. In this case, series \(\sum a_{n}\) will also converge to A, because the zero terms have no impact on the limit.

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