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Question: Test the alternating series \(\sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} \) from n is \([2,\infty )\) fo...

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

Explanation

Solution

Given a series ∑an=∑(−1)nln⁡n\sum {{a_n} = } \sum {\dfrac{{{{( - 1)}^n}}}{{\ln n}}} . Now apply the alternating series.
Consider the sequencebn=1ln⁡n{b_n} = \dfrac{1}{{\ln n}} , since logarithmic function is the positive function in [2,∞)[2,\infty )therefore, bn⩾0{b_n} \geqslant 0 all n.
We have to prove bn=1ln⁡n{b_n} = \dfrac{1}{{\ln n}} is decreasing. For this we have to prove bn+1<bn{b_{n + 1}} < {b_n}.
Now, prove lim⁡n→∞bn=0\mathop {\lim }\limits_{n \to \infty } {b_n} = 0
Use the fact 1ln⁡∞=0\dfrac{1}{{\ln \infty }} = 0

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

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

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

Note: Alternating series test:
Suppose that we have a series ∑an\sum {{a_n}} and either an=(−1)nbn{a_n} = {( - 1)^n}{b_n} or an=(−1)n+1bn{a_n} = {( - 1)^{n + 1}}{b_n} where bn⩾0{b_n} \geqslant 0 all n. Then if,
lim⁡n→∞bn=0\mathop {\lim }\limits_{n \to \infty } {b_n} = 0
and
bn\\{ {b_n}\\} is a decreasing sequence
the series ∑an\sum {{a_n}} is convergent.