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. \(^{n \rightarrow \infty}\) $$\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$

Short answer

Answer: The given telescoping series converges to \(\frac{1}{2}\).

Step-by-step solution

Find the nth term of the sequence of partial sums

To do this, we need to rewrite the given series as a sum of partial sums: $$S_n = \sum_{k=1}^{n}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$$ To find the simplified series, let's write the first few terms and try to cancel out terms that appear more than once: $$S_n = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \left(\frac{1}{4}-\frac{1}{5}\right)+ \cdots+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$$ As we can see, there will be a lot of cancellation of the terms. The simplified partial sum is given by: $$S_n= \frac{1}{2}-\frac{1}{n+2}$$

Evaluate the limit of the sequence of partial sums as n approaches infinity

Now, let's find the limit of the sequence of simplified partial sums as n approaches infinity: $$\lim_{n\to\infty}S_n = \lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{n+2}\right)$$ The second term \(\frac{1}{n+2}\) approaches 0 as n becomes infinitely large. Thus, the limit is: $$\lim_{n\to\infty}S_n = \lim_{n\to\infty}\frac{1}{2}-0 = \frac{1}{2}$$ So, the value of the series converges and is equal to \(\frac{1}{2}\)

Key concepts

Sequence of Partial Sums

When dealing with a telescoping series, like the one in the exercise, a sequence of partial sums helps us understand how the terms accumulate to form the series. Each term in the sequence of partial sums, denoted as \(S_n\), represents the sum of the first \(n\) terms of the series.

The series given in the exercise is \(\egin{equation*}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\d{equation*}\). Let's break it down: for each positive integer \(k\), the series adds \(\frac{1}{k+1}\) and subtracts \(\frac{1}{k+2}\). This pattern is key to the telescoping effect, where many terms cancel out each other.

To find the \(n\)th partial sum, we examine the pattern of cancellation within the series:- The first term is \(S_1 = \frac{1}{2} - \frac{1}{3}\)
- The second term is added: \(\frac{1}{3} - \frac{1}{4}\)
- Successive terms cancel previous terms, leaving: \(\frac{1}{2} - \frac{1}{n+2}\)
This simplification makes evaluating the series much more manageable.

Convergence of Series

A series converges if the sequence of its partial sums approaches a specific finite number as \(n\) goes to infinity. Understanding convergence is essential to determine whether a series has a sum or diverges to infinity or some undefined value.

In this telescoping series, we have already derived the \(n\)th partial sum:\[S_n = \frac{1}{2} - \frac{1}{n+2}\]By examining \(S_n\), we can state that as \(n\) tends towards infinity, \(\frac{1}{n+2}\) shrinks to 0. This happens because the denominator \(n+2\) becomes infinitely large.

Thus, the partial sums sequence approaches:- A single fixed value, \(\frac{1}{2}\), and we say the series converges to \(\frac{1}{2}\).

Failure to account for convergence could lead to confusing situations where a series seems to extend infinitely without any bound, not providing any significant insight into its ultimate behavior.

Limit of a Sequence

The limit of a sequence involves finding the value that the terms of the sequence approach as the index of the terms becomes very large. In mathematical terms, it’s about identifying what the terms look like at infinity.

In our exercise, we found the \(n\)th partial sums which are:\[S_n = \frac{1}{2} - \frac{1}{n+2}\]As \(n\) increases, the term \(\frac{1}{n+2}\) approaches zero. Why? Because increasing \(n\) makes \(n+2\) larger, causing its reciprocal to shrink towards zero. This is a classical example of calculating the limit of a sequence as \(n o \infty\).

Through calculation:- \(\lim_{n \to \infty} S_n = \frac{1}{2} - 0 = \frac{1}{2}\)
This limit tells us that the infinite series has a sum equal to \(\frac{1}{2}\). Recognizing and calculating limits are powerful tools in understanding infinite processes in calculus. The concept of limits helps us deal with infinity in a concrete way.