Question

For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\\{S_{n}\right\\} .\) Then evaluate lim \(S_{n}\) to obtain the value of the series or state that the series diverges. \(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty} \frac{1}{(a k+1)(a k+a+1)},$$ where \(a\) is a positive integer

Short answer

$$\lim_{n \to \infty} \sum_{k=1}^{\infty} \frac{1}{(a k+1)(a k+a+1)}$$ Answer: The limit of the given telescoping series as n approaches infinity is: $$\frac{1}{2(a+1)}$$

Step-by-step solution

Identify the series

The given telescoping series is: $$\sum_{k=1}^{\infty} \frac{1}{(a k+1)(a k+a+1)}$$

Simplify the series

We begin by decomposing the given summand into partial fractions, which will help us find a formula for the nth term of the partial sum: $$\frac{1}{(a k+1)(a k+a+1)}=\frac{A}{a k+1}+\frac{B}{a k+a+1}$$ Now, we need to find the values of A and B.

Determine values of A and B

Clear the denominators out: $$1 = A(a k + a + 1) + B(a k + 1)$$ Now, we can try plugging in some convenient values to find A and B. For example, let \(k=-1\) and \(k=0\). For \(k=-1\): $$1 = A(a(-1) + a + 1) = A(2) \Rightarrow A=\frac{1}{2}$$ For \(k=0\): $$1 = B(a(0) + 1) = B \Rightarrow B=\frac{1}{2}$$

Rewrite the summand with A and B

Now we can rewrite the summand: $$\frac{1}{(a k+1)(a k+a+1)}=\frac{1/2}{a k+1}+\frac{1/2}{a k+a+1}$$

Define the sequence of partial sums

The sequence of partial sums, \(S_n\), is given by: $$S_n = \sum_{k=1}^n \frac{1}{2}\left(\frac{1}{a k+1}-\frac{1}{a k+a+1}\right)$$

Simplify the partial sums and find a formula for the nth term

First, we notice how the series cancel out when adding the terms. For example, with the second term, the series becomes: $$S_2=\frac{1}{2}\left(\frac{1}{a+1}-\frac{1}{2a+2}\right)+\frac{1}{2}\left(\frac{1}{2a+2}-\frac{1}{3a+3}\right)$$ Notice that most of the terms cancel out. This pattern will continue and we can consolidate the remaining terms into: $$S_n = \frac{1}{2}\left(\frac{1}{a+1}-\frac{1}{na+a+1}\right)$$

Find the limit of the series as n approaches infinity

Now, we'll take the limit as n approaches infinity: $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{2}\left(\frac{1}{a+1}-\frac{1}{na+a+1}\right)$$ Since \(1/(na+a+1)\) approaches 0 as n approaches infinity, $$\lim_{n \to \infty} S_n= \frac{1}{2}\left(\frac{1}{a+1}\right)$$ Therefore, the value of the series is: $$\boxed{\frac{1}{2(a+1)}}$$

Key concepts

Partial Fractions

In calculus, particularly when working with series, breaking down complex fractions into simpler ones using partial fractions is a key technique. This methodology helps in finding trackable patterns within the series. The aim of partial fractions is to express a fraction as a sum of simpler fractions, making it easier to handle mathematically.

To decompose a fraction using partial fractions, we set up an equation where the complex fraction is equal to a sum of simpler fractions with unknown coefficients. For example, given \[ \frac{1}{(a k+1)(a k+a+1)} \]it can be expressed as:

• \( \frac{A}{a k+1} + \frac{B}{a k+a+1} \)

In this situation, \(A\) and \(B\) are coefficients to be determined. By equating the original fraction with this setup and substituting suitable values for \(k\), we can solve for \(A\) and \(B\). For this exercise, both coefficients came out to be \(\frac{1}{2}\). This simplification significantly eases finding the formula for the nth term of the partial sums.

Sequence of Partial Sums

A sequence of partial sums is incredibly helpful when you need to analyze the overall behavior of a series. Each term in this sequence represents the sum of the first \(n\) terms of the series, denoted by \(S_n\).

• Start from \(S_1\), the first term alone.
• \(S_2 = S_1\) plus the second term, and so on.

For this telescoping series, the sequence of partial sums is structured as:\[ S_n = \sum_{k=1}^n \frac{1}{2}\left(\frac{1}{a k+1} - \frac{1}{a k+a+1}\right) \]Interestingly, the terms within this series cancel each other significantly. For instance:

• \(S_2 = \frac{1}{2}\left(\frac{1}{a+1} - \frac{1}{2a+2}\right) + \frac{1}{2}\left(\frac{1}{2a+2} - \frac{1}{3a+3}\right)\), here, the middle terms cancel out.

This pattern leads to a telescoping effect where only a few terms remain which reveals the nature of the series, making it significantly easier to assess its convergence.

Limit of a Series

Determining whether a series converges or diverges is often dependent on evaluating the limit of its sequence of partial sums. If \( \lim_{n \to \infty} S_n \) exists and is finite, the series converges.

For this specific telescoping series:\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{2}\left(\frac{1}{a+1} - \frac{1}{na+a+1}\right) \]As \(n\) approaches infinity, the term \(\frac{1}{na+a+1}\) approaches zero, allowing the limit to converge to:\[ \frac{1}{2}\left(\frac{1}{a+1}\right) \]This result confirms that the series converges to the finite sum \( \frac{1}{2(a+1)} \). Understanding this concept is fundamental for analyzing the behavior and outcome of infinite series.

Series Divergence

When studying series, it's crucial to know when they diverge, meaning they do not settle into a specific number as they expand to infinity. If the limit of the sequence of partial sums, \( \lim_{n \to \infty} S_n \), doesn't exist or is infinite, then the series is divergent.

In our worked example, this was not the case; however, knowing how to identify divergence is key when handling infinite series in general.

Here are some signs that may indicate divergence:

• If \(S_n\) becomes unbounded as \(n\) increases, the series diverges.
• If the terms within the series do not tend to zero as \(n\) increases, the series diverges. For a series to converge, the individual terms should approach zero.

Recognizing divergence early can save a lot of time in addressing how to properly evaluate or approximate series.