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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.2.48

Find the sum of each series in Exercises 45–52.
∑ (from n = 1 to ∞) [ (2n + 1) / (n²(n + 1)²) ]

Verified step by step guidance
1
Start by examining the general term of the series: \(\frac{2n + 1}{n^{2}(n + 1)^{2}}\). Our goal is to express this term in a form that allows us to use telescoping or known series sums.
Try to decompose the term into partial fractions. Since the denominator is \(n^{2}(n + 1)^{2}\), consider expressing the term as a sum of fractions of the form \(\frac{A}{n} + \frac{B}{n^{2}} + \frac{C}{n + 1} + \frac{D}{(n + 1)^{2}}\).
Set up the equation: \(\frac{2n + 1}{n^{2}(n + 1)^{2}} = \frac{A}{n} + \frac{B}{n^{2}} + \frac{C}{n + 1} + \frac{D}{(n + 1)^{2}}\). Multiply both sides by \(n^{2}(n + 1)^{2}\) to clear denominators and solve for constants \(A\), \(B\), \(C\), and \(D\) by equating coefficients of powers of \(n\).
Once the partial fractions are found, rewrite the series as a sum of simpler series: \(\sum_{n=1}^{\infty} \left( \frac{A}{n} + \frac{B}{n^{2}} + \frac{C}{n + 1} + \frac{D}{(n + 1)^{2}} \right)\). This will allow you to separate the original series into sums that are easier to evaluate or telescope.
Evaluate the sums by recognizing telescoping patterns or using known series results such as the Riemann zeta function for \(\sum \frac{1}{n^{2}}\). Combine the results carefully to find the sum of the original series.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. To find its sum, the series must converge, meaning its partial sums approach a finite limit. Understanding convergence tests helps determine if the series sum exists.
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