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. \(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty}\left(\frac{1}{\sqrt{k+1}}-\frac{1}{\sqrt{k+3}}\right)$$

Short answer

Answer: The value of the convergent telescoping series is \(\frac{1}{\sqrt{2}}\).

Step-by-step solution

Write down the general term of the series

The general term of the given telescoping series is given by: $$a_k = \frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}}.$$

Write down the partial sum of the series up to n terms

Now let's find the partial sum of the series up to n terms. The nth partial sum, \(S_n\), is: $$S_n = \sum_{k=1}^{n}\left(\frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}}\right).$$

Cancel the terms to find a simplified expression for partial sum

We'll now cancel the terms to get a simplified expression for the partial sum: $$S_n = \left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{4}}\right) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}}\right) + \cdots + \left(\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+3}}\right).$$ Notice that most of the terms cancel out, resulting in a simplification: $$S_n = \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{n+3}}.$$

Find the limit of the partial sum as n approaches infinity

We now evaluate the limit of the partial sum \(S_n\) as \(n \rightarrow \infty\) to determine whether the series converges or diverges: $$\lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n+3}}\right).$$ As \(n\) goes to infinity, the term \(\frac{1}{\sqrt{n+3}}\) goes to 0. Thus, we get the limit: $$\lim_{n \rightarrow \infty} S_n = \frac{1}{\sqrt{2}}.$$ Since the limit exists and it is a finite number, the series converges, and the value of the series is: $$\sum_{k=1}^{\infty}\left(\frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}}\right) = \frac{1}{\sqrt{2}}.$$

Key concepts

Convergence of Series

When we talk about a series converging, we mean that as we keep adding more and more terms, the total sum gets closer and closer to a certain number. This number is called the limit of the series. In the case of a telescoping series, like the one presented in the exercise, terms cancel out, simplifying the process of finding this limit. You are mostly left with the first and last few terms after all the cancellations. For this series, after canceling out the unnecessary terms, we ended up with the partial sum being equal to \( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n+3}} \). As you can see, it's much easier to find the limit since it boils down to observing the behavior of \( \frac{1}{\sqrt{n+3}} \) as \( n \rightarrow \infty \). As \( n \) becomes infinitely large, the \( \frac{1}{\sqrt{n+3}} \) term approaches zero, meaning the sum converges to \( \frac{1}{\sqrt{2}} \). Thus, the series converges and this finite limit is the value at which it converges.

Partial Sum

The partial sum of a series is simply the sum of the first few terms of a series, up to the nth term. It is represented as \( S_n \). Finding the partial sum is an essential step in determining the overall behavior of a series. It allows us to evaluate if the series is convergent or divergent. In the exercise, we derived the nth partial sum for the telescoping series: \[ S_n = \sum_{k=1}^{n}\left(\frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}}\right). \]We perform telescoping by simplification through cancelation of terms. Almost all terms cancel, leaving a much simpler expression. That's how we ended with:\[ S_n = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n+3}}. \]This step is crucial because once simplified, it's easier to determine the limit as \( n \) approaches infinity. If \( S_n \) gets closer to a certain number, then the series converges to that number.

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus that deals with the behavior of a sequence as the index, usually called \( n \), approaches infinity. For a series to converge, its sequence of partial sums must have a finite limit. Essentially, this means as you add more terms, the value you end up with should start to stay around a certain number.In our telescoping series exercise, we found this limit by evaluating the following:\[ \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n+3}}\right). \]The term \( \frac{1}{\sqrt{n+3}} \) becomes insignificantly small (approaches zero) as \( n \) tends toward infinity. So effectively, the expression left is \( \frac{1}{\sqrt{2}} \).Through this process, we concluded that our series converges to \( \frac{1}{\sqrt{2}} \) since the limit of the sequence of partial sums exists and is finite.

Divergent Series

A series is considered to be divergent if the total sum does not approach any limit as more terms are added. This means that instead of settling around a specific value, a divergent series grows without bound or oscillates indefinitely.In the context of the exercise, the series does not diverge, because we found that the limit as \( n \rightarrow \infty \) is a finite value \( \frac{1}{\sqrt{2}} \). However, understanding what a divergent series entails is important:- The series could increase or decrease without approaching a fixed number.- Divergent series are vital in understanding where and why certain series fail to behave predictably. They remain limitless.For instance, the harmonic series \( \sum_{k=1}^{ \infty \frac{1}{k}} \) is a classic example of a divergent series because its partial sums grow infinitely large.In summary, while the telescoping series in our problem converges, knowing about divergent series helps in recognizing cases where series do not have finite limits.