Question

Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ c. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series.

Short answer

The given series converges with a sum of \(1/2\).

Step-by-step solution

Apply Convergence Test

To analyze the convergence of the series, one can apply the Ratio Test, having that the terms are all positive. Therefore, a ratio between \(a_{n+1}/a_n\) must be calculated where \(a_n\) is the nth term of the series. This ratio tends to 1 as \(n\) approaches infinity, which means the series converges due to the ratio test.

Verify the Series Equality

In this step, the series equality is verified by performing a fractional decomposition of \(\frac{1}{(n+2)(n+1)}\), which results in \(frac{n+1}{n+2}-\frac{n}{n+1}\) as given. This confirms the equality.

Find the Nth Partial Sum

The nth partial sum of the series can be computed by summing up the first ‘n’ terms of the series. Since this is a telescoping series, most terms cancel out, and the partial sum \(\frac{n+1}{n+2}-\frac{1}{2}\) remains.

Determine the Sum of the Telescoping Series

The sum of the telescoping series can be calculated by taking the limit of the nth partial sum as \(n\) approaches infinity, which results in \(1-1/2 = 1/2\).

Key concepts

Convergence Test

When dealing with an infinite series like \[\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)},\]it's important to first verify whether the series converges or diverges. One of the methods to perform this check is using the **Ratio Test**. The Ratio Test helps us determine if an infinite series converges by comparing the ratio of consecutive terms.For a series \( \sum a_n \), the Ratio Test involves computing the limit:\[L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}.\]- If \( L < 1 \), the series converges.- If \( L > 1 \) or \( L \) is infinite, the series diverges.- If \( L = 1 \), the test is inconclusive.In this specific series \[\frac{1}{(n+2)(n+1)},\]applying the Ratio Test shows that \( L = 1 \), which initially suggests the test is inconclusive. However, care should be taken to apply further steps or simplifications to verify convergence properly. Once other criteria are applied, such as looking for simpler expressions through decomposition, convergence can be affirmed, as explored in the later sections.

Partial Sum

A **partial sum** of a series \( \sum a_n \) up to \( n \) terms is simply the sum of the first \( n \) terms. It is expressed as:\[S_n = \sum_{k=1}^{n} a_k.\]For our given series, after decomposing \( \frac{1}{(n+2)(n+1)} \) into the form \( \frac{n+1}{n+2} - \frac{n}{n+1} \), calculating the partial sum becomes quite straightforward due to the nature of these terms.Because many terms in the series cancel each other out, it's much easier to manage. Specifically, each term cancels with a part of the subsequent term, leading to a much simpler expression for the partial sum:\[S_n = \frac{1}{2} - \frac{1}{n+2}.\]Using this approach simplifies calculations and helps to establish the series' behavior as \( n \to \infty \). You'll see in a telescoping series, this cancellation becomes particularly handy.

Telescoping Series

A **telescoping series** is a special kind of series where many of the terms cancel when summed. This makes finding the total sum or a specific partial sum much simpler.The given series, \[\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2} - \frac{n}{n+1}\right),\]is an example of a telescoping series. When you expand the terms, you'll observe:

• The term \( \frac{n+1}{n+2} \) partially cancels with \( -\frac{n+1}{n+2} \) from the next term.
• As you progress through larger \( n \), these cancellations simplify the series significantly.

Only the first few terms and the last surviving terms matter in a partial sum, leading to shorter calculations. Specifically, the partial sum formula\[S_n = \frac{1}{2} - \frac{1}{n+2}\]emerges from this telescoping pattern. This wonderful feature makes it easy to grasp convergence properties and results.

Ratio Test

The **Ratio Test** is a handy tool in understanding the behavior of infinite series. This test compares consecutive terms of the series to evaluate its convergence.Apply it by:

• Calculating the ratio: \( \frac{a_{n+1}}{a_n} \) for the given series terms.
• Taking the limit, \( L \), as \( n \) approaches infinity.

For our given series \[\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)},\]performing a ratio test initially suggests inconclusive results since:\[L = \lim_{n \to \infty} \frac{\frac{1}{(n+3)(n+2)}}{\frac{1}{(n+2)(n+1)}} = 1.\]This outcome where \( L = 1 \) means further analysis or refining methods (aside from the basic ratio test), like transformations or decompositions, may be needed to thoroughly assess convergence. Nevertheless, a thoughtful combination of different tests and transformations often leads to enlightening conclusions about the series' convergence or divergence.