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Calculus / Calculus: Early Transcendentals 1 / Chapter 8.3 / Problem 56E

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

Telescoping series For the following

Chapter 11, Problem 56E

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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.

$\sum_{k=0}^{\infty}\left[\sin \left(\frac{(k+1) \pi}{2 k+1}\right)-\sin \left(\frac{k \pi}{2 k-1}\right)\right]$

Questions & Answers

QUESTION:

$\sum_{k=0}^{\infty}\left[\sin \left(\frac{(k+1) \pi}{2 k+1}\right)-\sin \left(\frac{k \pi}{2 k-1}\right)\right]$

ANSWER:

Problem 56E

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 state that the series diverges.

$\sum\limits_{k\ =\ 0}^{\infty{}}[sin\ (\frac{(k\ +\ 1)\pi{}}{2k\ +\ 1})-sin\ (\frac{k\ \pi{}}{2k\ -1})]$

Solution

Step 1

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

Consider $\sum\limits_{k\ =\ 0}^{\infty{}}[sin\ (\frac{(k\ +\ 1)\pi{}}{2k\ +\ 1})-sin\ (\frac{k\ \pi{}}{2k\ -1})]$

Let us first find the ${n}^{th}$ term of the sequence of partial sums ${S}_{n}$ .

${S}_{n}=\ \sum\limits_{k\ =\ 0}^{n}[sin\ (\frac{(k\ +\ 1)\pi{}}{2k\ +\ 1})-sin\ (\frac{k\ \pi{}}{2k\ -1})]$ … (1)

Substitute values for $k=0,1,2,...,(n-1),n$ we get

${S}_{n}=[sin\ (\pi{})-sin\ (0)]+[sin\ (\frac{2\pi{}}{3})-sin\ (\pi{})]+[sin\ (\frac{3\pi{}}{5})-sin\ (\frac{2\pi{}}{3})]+...\ +[sin\ (\frac{n\pi{}}{2n-1})-sin\ (\frac{(n\ -\ 1)\pi{}}{2n\ -\ 3})]$

$+[sin\ (\frac{(n\ +\ 1)\pi{}}{2n\ +\ 1})-sin\ (\frac{n\pi{}}{2n\ -\ 1})]$

Cancelling the like terms with opposite sign we get,

${S}_{n}=\ -sin\ (0)+sin\ (\frac{(n\ +\ 1)\pi{}}{2n\ +\ 1})$

${S}_{n}=sin\ (\frac{(n\ +\ 1)\pi{}}{2n\ +\ 1})$ (Since sin 0 = 0)

Thus the ${n}^{th}$ term in the series is ${S}_{n}=sin\ (\frac{(n\ +\ 1)\pi{}}{2n\ +\ 1})$

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Telescoping series For the following

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Solution Found!

Telescoping series For the following

Questions & Answers

Study Tools You Might Need

Chapter: 9 Problem: 17P

Chapter: 3 Problem: Q8

Chapter: 6 Problem: 8P

Chapter: 5 Problem: 5-11

Chapter: 25 Problem: 25.31

Chapter: 13 Problem: 10

Chapter: 2 Problem: 1

Chapter: 19 Problem: 22

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