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Question

Question: How do you show the series \(\dfrac{1}{3} - \dfrac{2}{5} + \dfrac{3}{7} - \dfrac{4}{9} + ... + \dfra...

How do you show the series 1325+3749+...+(1)n1n2n+1+...\dfrac{1}{3} - \dfrac{2}{5} + \dfrac{3}{7} - \dfrac{4}{9} + ... + \dfrac{{{{( - 1)}^{n - 1}}n}}{{2n + 1}} + ... diverges?

Explanation

Solution

The given series is an alternating series, to show the series diverges and use Leibnitz test for alternating series. In the Leibnitz test, we perform two steps in which the first one we check whether the series is increasing or decreasing by comparing two consecutive terms of the series. And in the second step we find the “nth” term of the series where the limit of “n” tends to infinity is equals to zero or not, if it equals to zero then series is converging otherwise diverging.

Complete step by step solution:
In order to show the series 1325+3749+...+(1)n1n2n+1+...\dfrac{1}{3} - \dfrac{2}{5} + \dfrac{3}{7} - \dfrac{4}{9} + ... + \dfrac{{{{( - 1)}^{n - 1}}n}}{{2n + 1}} + ... diverges, we will use Leibnitz test for alternating series. Because the given series is an alternating series.
Now, performing the Leibnitz test as follows
First checking whether the series is increasing or decreasing:
If un>un+1\left| {{u_n}} \right| > \left| {{u_{n + 1}}} \right|
Letting the first and second term for this test, that is 13  and  25\dfrac{1}{3}\;{\text{and}}\; - \dfrac{2}{5}
13>25\left| {\dfrac{1}{3}} \right| > \left| { - \dfrac{2}{5}} \right| it does not hold good, that means the given series is an increasing series.
Now, we will perform the second step:
We will check either the “nth” term of the series where limit of “n” is tends to infinity is equals to zero or not
=limnun =limn(1)n1n2n+1 =limnn2n+1  = {\lim _{n \to \infty }}\left| {{u_n}} \right| \\\ = {\lim _{n \to \infty }}\left| {\dfrac{{{{( - 1)}^{n - 1}}n}}{{2n + 1}}} \right| \\\ = {\lim _{n \to \infty }}\dfrac{n}{{2n + 1}} \\\
Diving the numerator and denominator with “n” to solve the limit,
=limn12+1n= {\lim _{n \to \infty }}\dfrac{1}{{2 + \dfrac{1}{n}}}
Now, putting the limit, we will get
=12= \dfrac{1}{2}

Since Leibnitz test says if limnun0{\lim _{n \to \infty }}\left| {{u_n}} \right| \ne 0 then series diverges, therefore the given series diverges.

Note:
When taking the terms to compare in order to check if the series is either increasing or decreasing, make sure you are taking two consecutive terms and also put modulus on both terms when comparing them.