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Chapter 7: Problem 2

Find the first five terms of the sequence of partial sums. $$ \frac{1}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\frac{3}{4 \cdot 5}+\frac{4}{5 \cdot 6}+\frac{5}{6 \cdot 7}+\cdots $$

Short Answer

Expert verified
The first five terms of the sequence are: 0.167, 0.25, 0.3, 0.333, 0.357

Step by step solution

01

Identify the pattern

Each term in the infinite series can be described as: \(\frac{n}{(n+1)(n+2)}\) , where \(n\) starts from 1. This common structure provides an insight into the pattern of the sequence.
02

Break down the fraction

We see that each term in the sequence is a fraction. We can break down each term in the following way: \(\frac{n}{(n+1)(n+2)}=\frac{1}{n+1} - \frac{1}{n+2}\)
03

Find the first five terms

Now, compute the first five terms by adding the terms according to the derived formula:- For \(n=1\), the first term is: \(\frac{1}{1+1} - \frac{1}{1+2} = 0.5 - 0.333 = 0.167\)- For \(n=2\), the sum of first two terms is: \(0.167 + (\frac{1}{2+1} - \frac{1}{2+2}) = 0.167 + 0.333 - 0.25 = 0.25\)- For \(n=3\), the sum of first three terms is: \(0.25 + (\frac{1}{3+1} - \frac{1}{3+2}) = 0.25 + 0.25 - 0.2 = 0.3\)- For \(n=4\), the sum of first four terms is: \(0.3 + (\frac{1}{4+1} - \frac{1}{4+2}) = 0.3 + 0.2 - 0.167 = 0.333\)- For \(n=5\), the sum of first five terms is: \(0.333 + (\frac{1}{5+1} - \frac{1}{5+2}) = 0.333 + 0.167 - 0.143 = 0.357\)
04

Write down the sequence

The sequence of the first five partial sums is: 0.167, 0.25, 0.3, 0.333, 0.357

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Most popular questions from this chapter

In Exercises 81-84, give an example of a sequence satisfying the condition or explain why no such sequence exists. (Examples are not unique.) A monotonically increasing sequence that converges to 10

(a) find the common ratio of the geometric series, \((b)\) write the function that gives the sum of the series, and (c) use a graphing utility to graph the function and the partial sums \(S_{3}\) and \(S_{5} .\) What do you notice? $$ 1-\frac{x}{2}+\frac{x^{2}}{4}-\frac{x^{3}}{8}+\cdots $$

(a) Show that \(\int_{1}^{n} \ln x d x<\ln (n !)\) for \(n \geq 2\). (b) Draw a graph similar to the one above that shows \(\ln (n !)<\int_{1}^{n+1} \ln x d x\) (c) Use the results of parts (a) and (b) to show that \(\frac{n^{n}}{e^{n-1}}1\) (d) Use the Squeeze Theorem for Sequences and the result of part (c) to show that \(\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\frac{1}{e}\) (e) Test the result of part (d) for \(n=20,50,\) and 100 .

Let \(\sum a_{n}\) be a convergent series, and let \(R_{N}=a_{N+1}+a_{N+2}+\cdots\) be the remainder of the series after the first \(N\) terms. Prove that \(\lim _{N \rightarrow \infty} R_{N}=0\).

Compute the first six terms of the sequence \(\left\\{a_{n}\right\\}=\left\\{\left(1+\frac{1}{n}\right)^{n}\right\\}\) If the sequence converges, find its limit.
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