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Calculus / Calculus: Early Transcendentals 1 / Chapter 8.3 / Problem 58E

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

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Answer: Telescoping series For the following telescoping

Chapter 11, Problem 58E

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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.

\(\sum_{k=1}^{\infty}\left[\tan ^{-1}(k+1)-\tan ^{-1} k\right]\)

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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.

\(\sum_{k=1}^{\infty}\left[\tan ^{-1}(k+1)-\tan ^{-1} k\right]\)

ANSWER:

Problem 58E

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. $\sum\limits_{k\ =\ 1}^{\infty{}}\ [ta{n}^{-1}\ (k\ +\ 1)-ta{n}^{-1}\ k]$

Solution

Step 1

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

Consider $\sum\limits_{k\ =\ 1}^{\infty{}}\ [ta{n}^{-1}\ (k\ +\ 1)-ta{n}^{-1}\ k]$

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

${S}_{n}=\sum\limits_{k\ =\ 1}^{n}\ [ta{n}^{-1}\ (k\ +\ 1)-ta{n}^{-1}\ k]$ … (1)

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

${S}_{n}=(ta{n}^{-1}\ 2\ -ta{n}^{-1}\ 1)+(ta{n}^{-1}\ 3\ -ta{n}^{-1}\ 2)+...+(ta{n}^{-1}\ n-ta{n}^{-1}\ (n-1))$

$+(ta{n}^{-1}\ (n+1)-ta{n}^{-1}\ n)\$

Cancelling the like terms with opposite sign we get,

${S}_{n}=\ -ta{n}^{-1}\ 1\ +ta{n}^{-1}\ (n+1)$

$=\ -\frac{\pi{}}{4}+ta{n}^{-1}\ (n\ +1)$

Thus the ${n}^{th}$ term in the series is ${S}_{n}\ =\ -\frac{\pi{}}{4}+ta{n}^{-1}\ (n\ +1)$

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