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Chapter 9: Problem 100

Remainder Let \(\Sigma a_{n}\) be a convergent series, and let \(R_{N}=a_{N+1}+a_{N+2}+\cdots\) be the remainder of the series after the first \(N\) terms. Prove that \(\lim _{N \rightarrow \infty} R_{N}=0\)

Short Answer

Expert verified
If \(\Sigma a_{n}\) is a convergent series and \(R_{N}=a_{N+1}+a_{N+2}+\cdots\) is the remainder of the series after the first \(N\) terms, then indeed \(\lim _{N\rightarrow \infty} R_{N}=0\) as proven by the property of convergent series and taking limits.

Step by step solution

01

Understand the problem

In the problem, \(\Sigma a_{n}\) is given as a convergent series and \(R_{N}=a_{N+1}+a_{N+2}+\cdots\) is defined as the remainder of the series after the first \(N\) terms. It's asked to prove that \(\lim _{N\rightarrow \infty} R_{N}=0\). We know that a series is said to be convergent if the series of its partial sums is convergent. That is, if \(S_{n} = a_{1} + a_{2} + ... + a_{n}\) and this sequence \(s_{n}\) is convergent, then the initial series is convergent.
02

Use the property of convergent series

According to the property of convergent series, since the series \(\Sigma a_{n}\) is given as a convergent series, the sequence of partial sums \(S_{n}\) tends to a limit as \(n\) approaches infinity. So, let the limit be 'S'. This means, \(\lim _{n\rightarrow \infty} S_{n} = S.\) Now, consider the partial sum till \(N+1\) terms. i.e., \(S_{N+1} = a_{1} + a_{2} + ... + a_{N} + a_{N+1} .\) As per the definition, \(R_N = S - S_{N}\). So when we apply the limit, we get \(\lim _{N\rightarrow \infty} R_{N}\) = \(\lim _{N\rightarrow \infty} S - S_{N}\).
03

Apply the limit

Let us now apply the limit. We get \(\lim _{N\rightarrow \infty} S - S_{N} = S - \lim _{N\rightarrow \infty} S_{N}\). Since \(S_{N}\) also tend to 'S' as \(N\) approaches infinity, \(\lim _{N\rightarrow \infty} S_{N} = S\). Thus, \(\lim _{N\rightarrow \infty} S - S_{N} = S - S = 0\).

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