Problem 40 - Thomas Calculus

Question

In Exercises \(39-44\) , find a formula for the \(n\) th partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum. $$\sum_{n=1}^{\infty}\left(\frac{3}{n^{2}}-\frac{3}{(n+1)^{2}}\right)$$

Step-by-Step Solution

Answer

The series converges, and its sum is 3.

Step 1: Recognize the Form of the Series

The series given is \( \sum_{n=1}^{\infty}\left(\frac{3}{n^{2}}-\frac{3}{(n+1)^{2}}\right) \). This is a telescoping series because each term is made up of a pair of fractions that will mostly cancel out with the next terms.

Step 2: Write the General Formula for the nth Partial Sum

A telescoping series will have most intermediate terms cancel. For the terms \( \frac{3}{n^2} - \frac{3}{(n+1)^2} \), write out the expression for the nth partial sum, \( S_n \), which sums from \( n=1 \) to \( n=N \):\[S_N = \sum_{n=1}^{N}\left(\frac{3}{n^2} - \frac{3}{(n+1)^2}\right)\].

Step 3: Calculate the Partial Sum

Most terms in the sequence will cancel with their predecessors from earlier terms. For example: \[S_N = \left(3\left( \frac{1}{1^2} - \frac{1}{2^2} \right) + 3\left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \ldots + 3\left( \frac{1}{N^2} - \frac{1}{(N+1)^2} \right)\right)\]This simplifies to:\[S_N = 3\left(1 - \frac{1}{(N+1)^2}\right)\]

Step 4: Determine the Convergence of the Series

To determine convergence, examine the limit of \( S_N \) as \( N \) approaches infinity:\[\lim_{N \to \infty} S_N = \lim_{N \to \infty} 3\left(1 - \frac{1}{(N+1)^2}\right) = 3(1) = 3\]. Since the limit is a finite number, the series converges.

Step 5: Find the Sum of the Series

Since the series converges and the limiting value of the nth partial sum is 3, the sum of the infinite series is 3.

Key Concepts

Partial Sums

In the context of series, partial sums play a crucial role in understanding the behavior of an infinite series. A partial sum is simply the sum of a certain number of terms from the start of the series. For instance, if we have a series \( a_1 + a_2 + a_3 + \ldots \), the nth partial sum \( S_n \) is defined as:

• \( S_n = a_1 + a_2 + a_3 + \ldots + a_n \)

In our exercise, we're looking at a telescoping series of the form \( \sum_{n=1}^{\infty} \left(\frac{3}{n^2} - \frac{3}{(n+1)^2}\right) \). For this series, each term effectively cancels out most terms from earlier in the sequence. So, the partial sum formula becomes a lot simpler:

• \( S_N = 3\left(1 - \frac{1}{(N+1)^2}\right) \)

This simplification occurs because each fraction in the sequence has a partner or successor that negates it, except for the initial term and the last terms that don’t completely cancel. Partial sums help us understand convergence by showing us the behavior of the series as we sum more terms.
This streamlined sum is handy for determining whether the series converges or diverges and, if convergent, what value it converges to.

Convergence and Divergence of Series

Understanding whether a series converges or diverges is key in analysis. A series converges if its sequence of partial sums tends to a limit as the number of terms increases indefinitely. For our telescoping series, the partial sum expression \( S_N = 3\left(1 - \frac{1}{(N+1)^2}\right) \) plays an essential role here.
To determine convergence, we evaluate the limit:

• \( \lim_{N \to \infty} S_N = \lim_{N \to \infty} 3\left(1 - \frac{1}{(N+1)^2}\right) \)
• As \( N \to \infty \), the term \( \frac{1}{(N+1)^2} \to 0 \)
• So the limit becomes 3

Since this process results in a finite limit, the series is said to converge. If the series had approached an infinite value or did not settle on any value, it would diverge.
Convergence tells us that adding more terms to our partial sum will eventually approximate a stable sum, while divergence indicates the opposite – the sum grows without bound. Thus, understanding this behavior helps decide whether a series is bounded and meaningful in applications.

Limit of a Sequence

The limit of a sequence is a powerful concept that helps us understand the long-term behavior of a series. In mathematical terms, if a sequence \( a_n \) has a limit \( L \), we can say that as \( n \) approaches infinity, \( a_n \) gets arbitrarily close to \( L \). This idea is central to talking about the convergence of a series.
In our example, the telescoping series gave us the partial sum \( S_N = 3\left(1 - \frac{1}{(N+1)^2}\right) \). To find the limit of this sequence of partial sums as \( N \to \infty \), we observe:

• The term \( \frac{1}{(N+1)^2} \) approaches zero as \( N \to \infty \)
• Hence, \( S_N \) approaches 3

This tells us that the limit of the sequence of partial sums is 3, signifying convergence to this specific value.
Sometimes, finding limits can be tricky, especially for more complex series, but with practice, it reveals itself as a straightforward tool for analyzing series behavior. It's like predicting where a slowly narrowing path will end, helping to discern convergence and the stability of functions and processes described by series.