Problem 40 - Calculus: One and Several Variables

Question

Speed of convergence) Find the least integer N for which the \(n\)th partial sum of the series differs from the sum of the series by less than 0.0001. $$\sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{k}$$

Step-by-Step Solution

Answer

The least integer \( N \) for which the \( n \)th partial sum of the series \( \sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{k} \) differs from the sum of the series by less than 0.0001 is \( N=17 \).

Step 1: Find the sum of the series

Using the formula for the sum of an infinite geometric series, we have: \[ S = \frac{a_1}{1-r} = \frac{1}{1-\frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3. \] The sum of the series is 3.

Step 2: Find the partial sum formula

The partial sum formula for a geometric series is given by: \[ S_n = \frac{a_1(1-r^n)}{1-r}. \] For our series, \(a_1 = 1\), and \(r = \frac{2}{3}\), so we have: \[ S_n = \frac{1(1-\left(\frac{2}{3}\right)^n)}{1-\frac{2}{3}}. \]

Step 3: Set up an inequality for the difference between the sum and the partial sum

We want to find the least integer N for which the difference between the sum and the partial sum is less than 0.0001. So, we set up the following inequality: \[ 0.0001 > |S - S_n| = 3 - \frac{1(1-\left(\frac{2}{3}\right)^n)}{1-\frac{2}{3}}. \] Simplify the inequality: \[ \frac{1}{10,000} > \frac{(1-\left(\frac{2}{3}\right)^n)}{\frac{1}{3}}. \] Now, cross-multiply and solve for n.

Step 4: Solve the inequality for n

Cross-multiply and simplify: \[ 3,000 > 1-\left(\frac{2}{3}\right)^n. \] Now, subtract 1 from both sides: \[ 2,999 > \left(\frac{2}{3}\right)^n. \] To isolate n, we'll take the natural logarithm of both sides: \[ \ln(2,999) > n \ln\left(\frac{2}{3}\right). \] Now, divide both sides by \(\ln\left(\frac{2}{3}\right)\), which is a negative number, so the inequality sign will flip: \[ \frac{\ln(2,999)}{\ln\left(\frac{2}{3}\right)} < n. \] Using a calculator, we obtain \[ n > 16.7909\dots \]

Step 5: Find the least integer N

Since n must be an integer and n > 16.7909, the smallest possible value for n is 17. So, the least integer N for which the nth partial sum of the series differs from the sum of the series by less than 0.0001 is N=17.

Key Concepts

Infinite Series

An infinite series is a sum of an infinite number of terms. Essentially, it is an endless sequence of additions. In mathematics, infinite series can be represented by notation like \( \sum_{k=0}^{\infty} a_k \), where \( a_k \) are the individual terms of the series starting at \( k=0 \).
The challenge with infinite series is that they don't always add up to finite values. Some series diverge, meaning they grow ever larger without bound. However, others converge, getting closer and closer to a specific finite number as more terms are added.

• In the provided exercise, the infinite series begins at \( k=0 \) with the first term \( a_1 = 1 \) and a common ratio \( r = \frac{2}{3} \).
• This is an example of a geometric series, as each term after the first is obtained by multiplying the previous term by a constant ratio \( r \).

The series in question converges to a finite sum since the absolute value of the ratio \( r < 1 \), following the formula for the sum of an infinite geometric series:\[ S = \frac{a_1}{1-r} \]In this case, the series converges to the sum of 3.

Partial Sum

A partial sum takes a portion of an infinite series and adds up only a finite number of terms. While an infinite series can extend indefinitely, a partial sum, denoted as \( S_n \), is calculated by adding the first \( n \) terms.

• The partial sum helps in understanding how a series behaves as more and more terms are included.
• In the exercise, the partial sum \( S_n \) of the series is found using the formula:\[ S_n = \frac{a_1(1-r^n)}{1-r} \]

The formula is particularly useful in determining how close the partial sum is to the actual sum of the infinite series as \( n \) increases.
This method helps in evaluating the speed of convergence by checking the difference between the partial sum and the infinite series' actual sum. As \( n \) becomes larger, \( r^n \) becomes smaller, making the partial sum approach the series sum ever more closely.

Convergence

Convergence in the realm of series refers to the behavior where the sum approaches a particular value as more terms are added. For a series to converge, the limit of its partial sums \( S_n \) as \( n \) approaches infinity must equal the series' sum.

• Geometric series with \( |r|<1 \) are known to converge, and this exercise underscores this point.
• The difference \(|S-S_n|\) is crucial, as it quantifies how close the partial sum \( S_n \) is to the total sum \( S \) of the infinite series.

The goal in the original exercise is to find the smallest integer \( N \) such that this difference \(|S-S_n|\) is less than an arbitrarily small number, in this case, 0.0001.
This process involves setting up and solving an inequality, showing that as \( n \) increases, the partial sum \( S_n \) closely approximates the sum \( S \), demonstrating convergence.