Chapter 1: Problem 6

For the following series, write formulas for the sequences $a_{n}, S_{n},$ and $R_{n},$ and find the limits of the sequences as $n \rightarrow \infty$ (if the limits exist). $$\sum_{1}^{\infty} \frac{1}{n(n+1)} \quad \text { Hint: } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

Short Answer

Expert verified

The sequences are: $ a_{n}=\frac{1}{n} - \frac{1}{n+1} $, $ S_{n}=1-\frac{1}{n+1} $, and $ R_{n} = \frac{1}{n+1} $. The limits are $ \lim_{{n \to \infty}} S_{n} = 1 $ and $ \lim_{{n \to \infty}} R_{n} = 0 $.

Step by step solution

- Write the General Term

Start by writing the general term of the given series. Given the hint: equation: $ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} $Therefore, $ a_{n} = \frac{1}{n} - \frac{1}{n+1} $.

- Write the Partial Sum

The partial sum $ S_{n} $ for the series is the sum of the first $ n $ terms of the series. Using the general term from Step 1, and noting the telescoping nature:equation: \[ S_{n} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) \]

- Evaluate the Partial Sum

Evaluate the partial sum by expanding the series:equation: \[ S_{n} = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \]Most terms cancel out, leaving:equation: $ S_{n} = 1 - \frac{1}{n+1} $

- Find the Limit of the Partial Sum

The next step is to find the limit of $ S_{n} $ as $ n \to \infty $:equation: \[ \lim_{{n \to \infty}} S_{n} = \lim_{{n \to \infty}} \left(1 - \frac{1}{n+1}\right) \]Since $ \frac{1}{n+1} \to 0 $ as $ n \to \infty $, the limit is:equation: $ \lim_{{n \to \infty}} S_{n} = 1 $.

- Write the Formula for the Remainder

The remainder term $ R_n $ can be written as the difference between the infinite sum and the partial sum $ S_n $:equation: $ R_{n} = 1 - S_{n} = \frac{1}{n+1} $.

- Find the Limit of the Remainder

The final step is to find the limit of the remainder $ R_{n} $ as $ n \to \infty $:equation: \[ \lim_{{n \to \infty}} R_{n} = \lim_{{n \to \infty}} \frac{1}{n+1} = 0 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Telescoping Series

A telescoping series is a series where most terms cancel out when summed, making it easier to find the sum. In our example, the series $\frac{1}{n(n+1)}$ can be rewritten as $ \frac{1}{n} - \frac{1}{n+1} $. This decomposition reveals its telescoping nature. When we sum the series, intermediate terms cancel each other, greatly simplifying the computation. This is evident from the partial sum calculation, which collapses into just a few terms.

Partial Sum

The partial sum, denoted as $ S_{n} $, is the sum of the first $ n $ terms of a series. For our telescoping series, the partial sum is \[ S_{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots \; + \frac{1}{n} - \frac{1}{n+1} \] Observing this summation, you’ll notice that all intermediate terms cancel out, leaving $ S_{n} = 1 - \frac{1}{n+1} $. This simplification is crucial to finding the series sum efficiently.

Limit of a Sequence

The limit of a sequence $ \lim_{n \to \infty} a_{n} $ is the value that $ a_{n} $ approaches as $ n $ becomes infinitely large. For our series, we found the partial sum limit: \[ \lim_{n \to \infty} S_{n} = \lim_{n \to \infty} $ 1 - \frac{1}{n+1}$ = 1 \] This shows that as $ n $ increases, the value of $ S_{n} $ gets closer and closer to 1.

Remainder Term

The remainder term $ R_n $ in an infinite series is the difference between the infinite sum and the partial sum. For our series, this can be written as \[ R_{n} = 1 - S_{n} = \frac{1}{n+1} \] This term gives us insight into how close our partial sums are to the actual sum of the series. By finding the limit of the remainder term, we see \[ \lim_{n \to \infty} R_{n} = \lim_{n \to \infty} \frac{1}{n+1} = 0 \]. Therefore, as $ n $ becomes very large, $ R_{n} $ approaches 0, confirming the accuracy of our partial sums.

Infinite Series

An infinite series is a sum of infinitely many terms. In our problem, the series is given by \[ \sum_{1}^{\infty} \frac{1}{n(n+1)}\] By breaking it down, we see it has a telescoping nature, which simplifies dealing with an infinite number of terms. The concept of convergence comes into play here—an infinite series converges if the sum of its terms approaches a finite value as the number of terms approaches infinity. Our series converges to 1, confirmed by evaluating the limit of the partial sums.

For the following series, write formulas for the sequences \(a_{n}, S_{n},\) and \(R_{n},\) and find the limits of the sequences as \(n \rightarrow \infty\) (if the limits exist). $$\sum_{1}^{\infty} \frac{1}{n(n+1)} \quad \text { Hint: } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$