41–44. {Use of Tech} Remainders and estimates Consider the following convergent series.


a. Find an upper bound for the remainder in terms of n.


43. ∑ (k = 1 to ∞) 1 / 3ᵏ

Verified step by step guidance

1. Recognize that the series \( \sum_{k=1}^{\infty} \frac{1}{3^k} \) is a geometric series with the first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \).
2. Recall that for a convergent geometric series with \( |r| < 1 \), the sum to infinity is \( S = \frac{a}{1 - r} \).
3. The remainder \( R_n \) after summing the first \( n \) terms is the difference between the total sum and the partial sum: \( R_n = S - S_n \).
4. The partial sum of the first \( n \) terms of a geometric series is \( S_n = a \frac{1 - r^n}{1 - r} \). Substitute \( a = \frac{1}{3} \) and \( r = \frac{1}{3} \) to express \( S_n \).
5. Express the remainder \( R_n \) in terms of \( n \) by using the formula \( R_n = S - S_n = \frac{a}{1 - r} - a \frac{1 - r^n}{1 - r} = a \frac{r^n}{1 - r} \). Substitute the values of \( a \) and \( r \) to get the upper bound for the remainder.

Key Concepts

Convergent Geometric Series

A geometric series is a sum of terms where each term is a constant multiple (common ratio) of the previous one. If the absolute value of the common ratio is less than 1, the series converges to a finite sum. For example, the series ∑ (1/3)^k converges because |1/3| < 1.

Remainder (Error) in Infinite Series

The remainder after n terms of a convergent series is the difference between the infinite sum and the partial sum up to n terms. It measures the error when approximating the infinite sum by a finite sum. Finding an upper bound for the remainder helps estimate how close the partial sum is to the total.

Formula for Remainder in Geometric Series

For a geometric series with first term a and common ratio r (|r|<1), the remainder after n terms is given by R_n = a * r^(n+1) / (1 - r). This formula provides an explicit upper bound for the error when approximating the infinite sum by the first n terms.