/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Let \(\sum a_{n}\) be a converge... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 116

Let \(\sum a_{n}\) 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
The proof utilizes the definition of convergence and expressions for the remainder term. By writing \(R_N\) in terms of the series' sum and the partial sum at \(N\), we can utilize the property of the convergent sequence to show every \(R_N\) lies between \(-\varepsilon\) and \(\varepsilon\), hence, its limit as \(N\) approaches infinity is 0.

Step by step solution

01

Understand the Definition of Convergent Series

A series is said to be convergent if the sequence of its partial sums \(S_n = \sum_{i=1}^{n} a_i\) converges to a finite limit. This means for every \(\varepsilon > 0\), there exists a number \(N \in \mathbb{N}\) such that for all \(n \geq N\), we have \(|S_n - S| < \varepsilon\), where \(S\) is the limit of the sequence of partial sums.
02

Write the Remainder \(R_N\) in terms of Partial Sums

By definition of the remainder of the series after the first \(N\) terms, we can express \(R_N\) as \(R_N = S - S_N\), where \(S\) is the sum of the full series. This is because \(R_N\) is basically what is left of the series when we subtract the sum of the first \(N\) terms from the total sum \(S\).
03

Apply the Definition of Convergent Series

According to the definition of a convergent series, for every \(\varepsilon > 0\), there exists a \(N\) such that for all \(n \geq N\), we have \(|S_n - S| < \varepsilon\). Another way to express this is that for all \(n \geq N\), we have \(-\varepsilon < S_n - S < \varepsilon\). Using \(R_N = S - S_N\), we can then write that \(-\varepsilon < S - S_n < \varepsilon\), which is equivalent to \(-\varepsilon < R_n < \varepsilon\). Hence, the limit as \(N\) approaches infinity of \(R_N\) is equal to 0.

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