(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Test the series for convergence or divergence

[tex]

\sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1} \frac{{\ln n}}{n}}

[/tex]

2. Relevant equations

If [tex]

b_{n + 1} \le b_n \,\& \,\mathop {\lim }\limits_{n \to \infty } b_n = 0

[/tex]

then the series is convergent.

3. The attempt at a solution

[tex]

b_n = \frac{{\ln n}}{n},\,\,\mathop {\lim }\limits_{n \to \infty } b_n = \mathop {\lim }\limits_{n \to \infty } \frac{{\ln n}}{n} = \frac{\infty }{\infty } = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {\ln n} \right)^\prime }}{{\left( n \right)^\prime }} = \mathop {\lim }\limits_{n \to \infty } \frac{{1/n}}{1} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0

[/tex]

so it passes the 2nd test.

[tex]

\begin{array}{l}

{\rm{when }}n = 1,\,\frac{{\ln n}}{n} = 0 \\

{\rm{when }}n = 2,\,\frac{{\ln 2}}{2} = 0.35 \\

\\

{\rm{when }}n = 1,\,b_n = 0,\,b_{n + 1} = 0.35\, > 0,\,b_{n + 1} > b_n \\

\end{array}

[/tex]

so it fails the 1st test, hence is divergent

But the example says:

[tex]

\begin{array}{l}

\left( {\frac{{\ln n}}{n}} \right)^\prime = \frac{{\left( {\ln n} \right)^\prime \left( n \right) - \left( {\ln n} \right)\left( n \right)^\prime }}{{\left( n \right)^2 }} = \frac{{\left( {1/n} \right)n - \left( {\ln n} \right)\left( 1 \right)}}{{n^2 }} = \frac{{1 - \ln n}}{{n^2 }} \\

\\

\frac{{1 - \ln x}}{{x^2 }} \le 0\,{\rm{whenever }}x \ge 1 \\

\end{array}

[/tex]

Therefore it passes both tests and is convergent.

But if I graph the derivative:

I get both positive and negative values past x=1, so is the example wrong when it says that when x>=1 that the derivitave is always <=0? Or did I make a mistake?

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# Homework Help: Alternating Series

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