Question

Test the alternating series $\sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} $ from n is $[2,\infty )$ for convergence?

Answer 1

Hint: Given a series $\sum {{a_n} = } \sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} $. Now apply the alternating series.
Consider the sequence${b_n} = \dfrac{1}{{\ln n}}$ , since logarithmic function is the positive function in $[2,\infty )$therefore, ${b_n} \geqslant 0$ all n.
We have to prove ${b_n} = \dfrac{1}{{\ln n}}$ is decreasing. For this we have to prove ${b_{n + 1}} < {b_n}$.
Now, prove $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$
Use the fact $\dfrac{1}{{\ln \infty }} = 0$

Complete step by step answer:
Consider the series is given as $\sum {{a_n} = } \sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} $. Let the sequence, ${b_n} = \dfrac{1}{{\ln n}}$.
Here , ${b_n} \geqslant 0$ all n as the logarithmic function is positive function in $[2,\infty )$.
According to the alternative series test, if $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$ , and \[\{ {b_n}\} \] is a decreasing sequence
Then the series$\sum {{a_n} = } \sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} $ is convergent.

We have to prove \[\{ {b_n}\} \] is a decreasing sequence, Take ${b_n} = \dfrac{1}{{\ln n}}$ then ${b_{n + 1}} = \dfrac{1}{{\ln (n + 1)}}$
Consider n+1>n for all n, and $\ln (n + 1) > \ln n$ for $n \in [2,\infty )$
Then, $\dfrac{1}{{\ln (n + 1)}} < \dfrac{1}{{\ln n}}$
$ \Rightarrow {b_{n + 1}} < {b_n}$
Since next term is less than the previous term therefore, \[\{ {b_n}\} \] is a decreasing sequence.

Next we have to find , $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$
$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\ln n}}$
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\ln n}} = \dfrac{1}{\infty }$
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\ln n}} = 0$
Hence, $\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$
From the alternating series test , $\sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} $ is convergent from n is $[2,\infty )$.

Note: Alternating series test:
Suppose that we have a series $\sum {{a_n}} $ and either ${a_n} = {( - 1)^n}{b_n}$ or ${a_n} = {( - 1)^{n + 1}}{b_n}$ where ${b_n} \geqslant 0$ all n. Then if,
$\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$
and
\[\{ {b_n}\} \] is a decreasing sequence
the series $\sum {{a_n}} $is convergent.