Solution Found!

Calculus / Calculus: Early Transcendentals 1 / Chapter 8.3 / Problem 57E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Change Question

Telescoping series For the following telescoping series,

Chapter 11, Problem 57E

(choose chapter or problem)

Get Unlimited Answers

QUESTION:

47-58. Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\{S_{n}\right\}\). Then evaluate \(\lim_{n\rightarrow\infty}\ S_n\), to obtain the value of the series or state that the series diverges.

\(\sum_{k=0}^{\infty} \frac{1}{16 k^{2}+8 k-3}\)

Questions & Answers

QUESTION:

47-58. Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums \(\left\{S_{n}\right\}\). Then evaluate \(\lim_{n\rightarrow\infty}\ S_n\), to obtain the value of the series or state that the series diverges.

\(\sum_{k=0}^{\infty} \frac{1}{16 k^{2}+8 k-3}\)

ANSWER:

Problem 57E

Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate $\lim\limits_{{n} \rightarrow {\infty{}}}\ {S}_{n}$ to obtain the value of the series or stale that the series diverges. $\sum\limits_{k\ =\ 0}^{\infty{}}\frac{1}{16{k}^{2}\ +\ 8k\ -3}$

Solution

Step 1

In this problem we have to find the formula for ${n}^{th}$ term ${S}_{n}$ in $\sum\limits_{k\ =\ 0}^{\infty{}}\frac{1}{16{k}^{2}\ +\ 8k\ -3}$ and then we have to evaluate $\lim\limits_{{n} \rightarrow {\infty{}}}{S}_{n}$ or we have state that the series diverges.

Consider $\sum\limits_{k\ =\ 0}^{\infty{}}\frac{1}{16{k}^{2}\ +\ 8k\ -3}$

First let us simplify the above expression.

Consider $\frac{1}{16{k}^{2}\ +\ 8k\ -3}=\frac{1}{(4k\ -\ 1)(4k\ +\ 3)}$ … (1)

We shall simplify the above expression by partial fractions.

$\frac{1}{(4k\ -\ 1)(4k\ +\ 3)}=\frac{A}{4k\ -\ 1}+\frac{B}{4k\ +\ 3}$ … (2)

$\frac{1}{(4k\ -\ 1)(4k\ +\ 3)}=\ \frac{A\ (4k\ +\ 3)\ +\ B(4k\ -\ 1)}{(4k\ -\ 1)\ (4k\ +\ 3)}$

Since the denominators are same, we can equate the numerator.

$1=A(4k\ +\ 3)+B(4k\ -\ 1)$

Put $k=\ -3/4$ we get,

$1=-4B$

$\Rightarrow{}B=\frac{-1}{4}$

Put $k=1/4$ we get

$1=4A$

$\Rightarrow{}A=\frac{1}{4}$

Substitute values of A and B in (2) we get

$\frac{1}{(4k\ -\ 1)(4k\ +\ 3)}=\frac{1}{4\ (4k\ -\ 1)}-\frac{1}{4\ (4k\ +\ 3)}$

Thus (1) becomes,

$\frac{1}{16{k}^{2}\ +\ 8k\ -3}=\frac{1}{4\ (4k\ -\ 1)}-\frac{1}{4\ (4k\ +\ 3)}$

$\sum\limits_{k\ =\ 0}^{\infty{}}\frac{1}{16{k}^{2}\ +\ 8k\ -3}=\sum\limits_{k\ =\ 0}^{\infty{}}\{\frac{1}{4\ (4k\ -\ 1)}-\frac{1}{4\ (4k\ +\ 3)}\}$

$=\sum\limits_{k\ =\ 0}^{\infty{}}\frac{1}{4}\{\frac{1}{(4k\ -\ 1)}-\frac{1}{(4k\ +\ 3)}\}$

Study Tools You Might Need

Chapter: 16 Problem: 4Q

Introductory Chemistry 5
What is an oxidizing agent? What is a reducing agent?

Chapter: 11 Problem: 74P

Physics with MasteringPhysics 4
Problem 74P A student on a piano stool rotates freely with an angular speed of 2.95 rev/s. The student holds a 1.25-kg mass in the outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.43 kg -m2, a value that re

Chapter: Problem: 4

Fundamentals of Physics 10
Express the following angles in radians: (a) 20.0, (b) 50.0, (c) 100. Convert the following angles to degrees: (d) 0.330 rad, (e) 2.10 rad, (f) 7.70 rad

Chapter: 6 Problem: 5

Microbiology: An Introduction 11
Assume you inoculated 100 facultatively anaerobic cells onto nutrient agar and incubated the plate aerobically. You then inoculated 100 cells of the same species onto nutrient agar and incubated the second plate anaerobically. Aer incubation for 24 hours, you should havea. more colonies on the ae

Chapter: 9 Problem: 9.4

Campbell Biology 9
During the redox reaction in glycolysis (step 6 in Figure 9.9), which molecule acts as the oxidizing agent? The reducing agent?

Chapter: 8 Problem: 16

Elementary Statistics: A Step by Step Approach 8th ed. 8
Soft Drink Consumption A researcher claims that the yearly consumption of soft drinks per person is 52 gallons. In a sample of 50 randomly selected people, the mean of the yearly consumption was 56.3 gallons. The standard deviation of the population is 3.5 gallons. Find the P-value for the test. On t

Chapter: 10 Problem: ap3.35

The Practice of Statistics 5
An investor is comparing two stocks, A and B. She wants to know if over the long run, there is a significant difference in the return on investment as measured by the percent increase or decrease in the price of the stock from its date of purchase. The investor takes a random sample of 50 annualized

Chapter: 8 Problem: 7

Behavior Modification: Principles and Procedures 6
What does the abbreviation CER stand for? ________

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

×

Register

Sign up for access to all content on our site!

Or login if you already have an account