Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series. $$ \sum_{k=2}^{\infty}\left(\frac{3}{(k-1)^{2}}-\frac{3}{k^{2}}\right) $$

Step-by-Step Solution

Answer

The series converges and its sum is 3.

Step 1: Recognize the Series Form

The series \( \sum_{k=2}^{\infty}\left(\frac{3}{(k-1)^{2}}-\frac{3}{k^{2}}\right) \) is a telescoping series. Telescoping series are recognized by having terms that cancel out with subsequent terms.

Step 2: Write the First Few Terms

For a telescoping series, it's helpful to write out the first few terms to see the cancellation pattern:\[\frac{3}{1^2} - \frac{3}{2^2}, \quad \frac{3}{2^2} - \frac{3}{3^2}, \quad \frac{3}{3^2} - \frac{3}{4^2}, \ldots\] Notice that each negative fraction cancels with part of the next positive fraction.

Step 3: Simplify the Sum of the Series

From step 2, the series simplifies as each successive term cancels out part of the previous term except for the first fraction of the initial term in the series:\[\left( \frac{3}{1^2} \right) + \left( -\frac{3}{2^2} + \frac{3}{2^2} \right) + \left( -\frac{3}{3^2} + \frac{3}{3^2} \right) + \ldots\]Thus, the sum is simply:\[\frac{3}{1^2} = 3\]

Step 4: Converge or Diverge

The series converges because the non-cancelling terms shrink in magnitude and eventually eliminate, leaving a fixed sum that does not grow past the initial terms.

Step 5: Write the Conclusion

Given that the series telescopes effectively to leave a single term, the series converges with a sum of 3. This can be confirmed by observing the pattern and continuation of cancelling terms.

Key Concepts

Convergence of Series

To understand convergence of series, think of it like stacking blocks that eventually stop growing in height. Convergence occurs when adding more and more terms in a sequence results in the sum approaching a particular limit, rather than growing infinitely.
In the context of the \(\sum_{k=2}^{\infty}\left(\frac{3}{(k-1)^{2}}-\frac{3}{k^{2}}\right)\) series, convergence is achieved because as we add more terms, the increments become negligible, allowing the series to settle at a fixed sum.
Converging series can be treated with various tests:

• Telescoping Series Test: This series falls perfectly into this category. Each term cancels the previous in a way that leaves a finite remainder. This assures that the entire sequence of additions ends up confined within set bounds.
• Comparison Test: We often compare with known converging series to infer behavior.

Specifically, in telescoping series, the cancellation ensures convergence by default, typically leading to a sum formed from a finite number of terms. Here, convergence means an eventual cessation of further growth in total value.

Sum of Series

Finding the sum of a series can sometimes feel like solving a puzzle. We need to see how the pieces fit together. For telescoping series, this process is more straightforward due to the natural cancellation of terms, leaving only a few terms whose sum we need to find.
Consider the given series: \(\sum_{k=2}^{\infty}\left(\frac{3}{(k-1)^{2}}-\frac{3}{k^{2}}\right)\). When expanded, the terms cancel neatly, leaving only the initial term.
To deduce the sum:

• Identify Remaining Terms: Write out initial few terms to visualize cancellation: \(\frac{3}{1^2} - \frac{3}{2^2}, \quad \frac{3}{2^2} - \frac{3}{3^2}, \ldots\)
• Simplify Where Possible: All negative terms cancel succeeding positives leaving only \(\frac{3}{1^2}\).

Thus, simply summing the non-cancelled parts gives us the series sum of 3. It underscores how telescoping series often reduce intricate fractions into basic additions.

Mathematical Patterns

Recognition of mathematical patterns plays a powerful role in simplifying complex series. In telescoping series, spotting the pattern of cancellation is key to effortlessly finding the solution. This pattern resembles dominos falling sequentially, one term knocking out the next.
In our series: \(\sum_{k=2}^{\infty}\left(\frac{3}{(k-1)^{2}}-\frac{3}{k^{2}}\right)\), recognizing the pattern involves noting:

• Each term \(\frac{3}{k^2}\) is negated by part of the next term, \(\frac{3}{(k+1)^2}\).
• This creates a chain reaction across the series that simplifies finding both convergence and sum.

Learning to distinguish such mathematical patterns not only aids in solving telescoping series but also enriches problem-solving in broader mathematical contexts. Therefore, capturing these patterns is more about observing relationships and sequences rather than detailed computation.