Problem 116 - Essential Calculus

Question

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 \(\lim _{N \rightarrow \infty} R_{N}=0\).

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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 \(S\).

Step 3: 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.