Question

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

Short answer

Question: Show that the remainder, $R_n$, in the sum of the infinite geometric series and the sum of its first n terms can be expressed as $R_n = \frac{r^{n}}{1-r}$. Answer: We have derived the expression for the remainder, $R_n$, as $R_n = \frac{r^{n}}{1-r}$.

Step-by-step solution

Understand the given expressions and relations

Recall that the sum of an infinite geometric series is given by \(S = \frac{1}{1-r}\), when the absolute value of common ratio \(r\) is less than 1. Also, the sum of the first \(n\) terms of the geometric series, denoted by \(S_n\), is given by the formula \(S_n = \frac{1-r^n}{1-r}\). According to the exercise, we need to find the error between these values, which is the remainder \(R_n\), so \(R_n = S - S_n\).

Express the difference between \(S\) and \(S_n\)

We are given that \(S = \frac{1}{1-r}\) and \(S_n = \frac{1-r^n}{1-r}\). To find the difference between them, we'll subtract \(S_n\) from \(S\). So, we have: $$ R_{n} = S - S_{n} = \frac{1}{1-r} - \frac{1-r^n}{1-r} $$

Simplify the expression

The next step is to simplify the expression for \(R_n\). To do this, we'll find a common denominator. Notice that the denominator of \(S\) is \((1-r)\), thus we'll multiply the numerator and denominator of the first term by (1-r) so that both fractions have the same denominator: $$ R_{n} = \frac{(1-r)(1)}{1-r} - \frac{1-r^n}{1-r} $$ Now that both fractions have the same denominator, we can combine them: $$ R_n = \frac{1 - r - (1 - r^n)}{1-r} $$

Simplify the numerator

Notice that the "1" terms in the numerator cancel each other. Therefore, we have: $$ R_n = \frac{-r + r^n}{1-r} $$

Factor out the r term

We can now factor out the r term in the numerator: $$ R_n = \frac{r(r^{n-1} - 1)}{1-r} $$

Observe the signs

Notice that the signs within the parenthesis on the numerator are opposite from our target expression. We can fix this by multiplying the top and the bottom of the fraction by -1: $$ R_n = \frac{-r(-r^{n-1} + 1)}{-1(1-r)} $$

Simplify the expression

Now, we can simplify the expression to obtain the desired formula for the remainder: $$ R_n = \frac{r^{n}}{1-r} $$ Thus, we have derived the given formula for the remainder \(R_n\).

Key concepts

Infinite Series

Infinite series are sums of infinitely many terms. They occur in a variety of math problems. Geometric series are a type of infinite series, defined by having each term as a constant ratio of the previous term. For instance, the series \( \sum_{k=0}^{\infty} r^{k} \) represents an infinite geometric series where \( r \) is the common ratio. A critical requirement is that \( |r| < 1 \) to ensure the series converges to a finite value. In layman's terms, this means the series will not explode to infinity but will settle on some number. In this infinite geometric series, the sum of all terms is calculated by the formula \( S = \frac{1}{1-r} \), reflecting the limiting value that the series approaches as more terms are added.

Sum of Series

Considering finite sums first helps understand infinite ones. The sum of the first \( n \) terms of a geometric series, known as a partial sum, is represented by \( S_n = \frac{1-r^n}{1-r} \). Why is this important? Because calculating an infinite sum directly is impractical. We approximate by summing a large number of terms.

• \( S_n \) gives us an approximate value of the total series by just calculating those first \( n \) terms.
• It captures how much of the infinite series we have accounted for.

As \( n \) increases, \( S_n \) becomes a better approximation of \( S \), inching closer and closer to the full sum described earlier.

Remainder Calculation

The remainder \( R_n \) tells us how much of the series is left out when approximating using \( S_n \). It's the difference between the infinite series \( S \) and the partial sum \( S_n \). This can be calculated using:\[ R_n = S - S_n = \frac{r^n}{1-r} \] Here's why it matters:

• Understanding \( R_n \) helps in knowing the error made when approximating an infinite series with a finite one.
• It quantifies how close \( S_n \) is to \( S \). Less remainder means a more accurate approximation.

This calculation showcases how each additional term in the series improves the estimation, getting us ever more precise in representing the sum of the infinite series.