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Found in: Page 443

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?$$\sum\limits_{n = 1}^\infty {\frac{1}{{\ln (n + 1)}}}$$

Convergent.

See the step by step solution

## Definition

A convergent series is an infinite sum of a sequence of numbers$$\sum\limits_{n = 1}^\infty {{a_n}}$$ that adds up to finite number.

A divergent series is a series that does not converge, so if add up all of the terms of the series$$\sum\limits_{n = 1}^\infty {{a_n}}$$ don't have finite sum.

## Find terms

Consider the series$$\sum\limits_{n = 1}^\infty {\frac{1}{{\ln (n + 1)}}}$$

When$$n = 1$$

\begin{aligned}\sum\limits_{n = 1}^\infty {\frac{1}{{\ln (n + 1)}}} &= \frac{1}{{\ln (1 + 1)}}\\ &= \frac{1}{{\ln (2)}}\\ &= 1.4427\end{aligned}

When$$n = 2$$,

\begin{aligned}{s_2} &= \sum\limits_{n = 1}^2 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}}\\ &= 1.4427 + 0.91023\\ &= 2.35293\end{aligned}

## find terms

When$$n = 3$$,

\begin{aligned}{s_3} &= \sum\limits_{n = 1}^3 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}}\\ &= 3.0743\end{aligned}

When$$n = 4$$,

\begin{aligned}{s_4} &= \sum\limits_{n = 1}^4 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}} + \frac{1}{{\ln (5)}}\\ &= 3.6956\end{aligned}

## Find terms

When$$n = 5$$,

\begin{aligned}{s_5} &= \sum\limits_{n = 1}^5 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}} + \frac{1}{{\ln (5)}} + \frac{1}{{\ln (6)}}\\ &= 4.2537\end{aligned}

When$$n = 6$$,

\begin{aligned}{s_6} &= \sum\limits_{n = 1}^6 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}} + \frac{1}{{\ln (5)}} + \frac{1}{{\ln (6)}} + \frac{1}{{\ln (7)}}\\ &= 4.7676\end{aligned}

## Find terms

When$$n = 7$$,

\begin{aligned}{s_7} &= \sum\limits_{n = 1}^7 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}} + \frac{1}{{\ln (5)}} + \frac{1}{{\ln (6)}} + \frac{1}{{\ln (7)}} + \frac{1}{{\ln (8)}}\\ &= 5.2485{\rm{ }}\\{\rm{When }}n = 8\\{s_8} &= \sum\limits_{n = 1}^8 {\frac{1}{{\ln (n + 1)}}} \\ &= \frac{1}{{\ln (2)}} + \frac{1}{{\ln (3)}} + \frac{1}{{\ln (4)}} + \frac{1}{{\ln (5)}} + \frac{1}{{\ln (6)}} + \frac{1}{{\ln (7)}} + \frac{1}{{\ln (8)}} + \frac{1}{{\ln (9)}}\\ &= 5.7036\end{aligned}

## Check convergence

Using result that if the sequence$$\left\{ {{s_n}} \right\}$$ is convergent, then $$\mathop {\lim }\limits_{n \to \infty } {s_n} = s$$exists is a real number.

So, the series $$\sum {{a_n}}$$is called convergent and the sequence $$\left\{ {{s_n}} \right\}$$is divergent, then the series is called divergent.

Let$${a_n} = \frac{1}{{\ln (n + 1)}}$$then

\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\ln (n + 1)}}\\ &= \frac{1}{{\ln (\infty )}}\\ &= \frac{1}{\infty }\\ &= 0{\rm{ (define) }}\end{aligned}

Therefore, the series is convergent.