Question

Find‌ ‌whether‌ ‌the‌ ‌following‌ ‌series‌ ‌are‌ ‌convergent‌ ‌or‌ ‌divergent.‌ ‌

.$1‌ ‌+‌ ‌\dfrac{1}{2}.\dfrac{{{x^2}}}{4}‌ ‌+‌ ‌\dfrac{{1.3.5}}{{2.4.6}}.\dfrac{{{x^4}}}{8}‌ ‌+‌ ‌\dfrac{{1.3.5.7.9}}{{2.4.6.8.10}}.\dfrac{{{x^6}}}{{12}}‌ ‌+‌ ‌...$.‌ ‌ ‌

Answer 1

Here we have to find the values of x for which the series converge or diverge.There are different tests there.we will start from d’Alembert Ratio Test and then Gauss Test.

Complete step-by-step answer:

Step 1: Let the given series be (to find proper )

$\sum {{u_n}} = \dfrac{1}{2}.\dfrac{{{x^2}}}{4} + \dfrac{{1.3.5}}{{2.4.6}}.\dfrac{{{x^4}}}{8} + \dfrac{{1.3.5.7.9}}{{2.4.6.8.10}}.\dfrac{{{x^6}}}{{12}} + ...$

Where ${u_n} = \dfrac{{1.3.5.7.9...(4n - 3)}}{{2.4.6.8.10...(4n - 2)}}.\dfrac{{{x^{2n}}}}{{4n}}\,\,\,,n \geqslant 1$

It is the $n^{th}$ term of the above sequence and we can get all terms of the sequence by putting different values of n.

Step 2:

$\dfrac{{{u_n}}}{{{u_{n + 1}}}} = \dfrac{{1.3.5.7.9...(4n - 3)}}{{2.4.6.8.10...(4n - 2)}}.\dfrac{{{x^{2n}}}}{{4n}}\, \times \dfrac{{2.4.6.8.10...(4n - 2)4n(4n + 2)}}{{1.3.5.7.9...(4n - 3)(4n - 1)(4n + 1)}}.\dfrac{{(4n + 4)}}{{{x^{2n + 2}}}}$

After cancelling out terms we get,

$= \dfrac{{4n(4n + 2)}}{{(4n - 1)(4n + 1)}} \times \dfrac{{(4n + 4)}}{{4n}} \times \dfrac{1}{{{x^2}}}$

$= \dfrac{{(4n + 2)(4n + 4)}}{{(4n - 1)(4n + 1)}}.\dfrac{1}{{{x^2}}}$

Step 3:

$\displaystyle \lim_{n \to \infty} \dfrac{u_n}{u_{n + 1}} = \displaystyle \lim_{n \to \infty} \dfrac{{(4n + 2)(4n + 4)}}{{(4n - 1)(4n + 1)}}.\dfrac{1}{x^2}$

We will take 4n common from each term of numerator and denominator and cancel that, then-

$= \displaystyle \lim_{n \to \infty} \dfrac{{(1 + \dfrac{1}{{2n}})(1 + \dfrac{1}{n})}}{{(1 - \dfrac{1}{{4n}})(1 + \dfrac{1}{{4n}})}}.\dfrac{1}{{{x^2}}}$

$= \dfrac{1}{{{x^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(as\,\,\displaystyle \lim_{n \to \infty} \dfrac{1}{n} = 0 )$

Step 4:Then using Ratio Test,$\sum {{u_n}}$ converges for

$\dfrac{1}{{{x^2}}} > 1$

$\Rightarrow {x^2} < 1$

$\Rightarrow {x^2} < 1$

$\Rightarrow |x| < 1$

$\Rightarrow - 1 < x < 1$

Step 5: Then using Ratio Test, diverges for

$\dfrac{1}{{{x^2}}} < 1$

$\Rightarrow {x^2} > 1$

$\Rightarrow x \in R - [ - 1,1]$

Step 6: Now we have to check for $x = \pm 1$. Let us try for Gauss’s Test.

Putting $x = \pm 1$ ,

$\dfrac{{{u_n}}}{{{u_{n + 1}}}}\,\,\,\, = \,\dfrac{{(4n + 2)(4n + 4)}}{{(4n - 1)(4n + 1)}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{(1 + \dfrac{1}{{2n}})(1 + \dfrac{1}{n})}}{{(1 - \dfrac{1}{{4n}})(1 + \dfrac{1}{{4n}})}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = (1 + \dfrac{3}{{2n}} + \dfrac{1}{{2{n^2}}}){(1 - \dfrac{1}{{16{n^2}}})^{ - 1}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \dfrac{3}{{2n}} + o(\dfrac{1}{{{n^2}}})$

Here the coefficient of $\dfrac{1}{n}\,\,is\,\,\dfrac{3}{2}\,\, > 1.$ Therefore by Gauss’s Test ,the series converges for $x \in [ - 1,1]$.

Then the series converges for $x \in [ - 1,1]$ and diverges for $x \in R - [ - 1,1]$.

Note: Here R is the set of real numbers. Also $o(\dfrac{1}{{{n^2}}})$ denotes the terms of $\dfrac{1}{n}$ having power greater than or equal to 2.