Question

Find whether the following series are convergent or divergent:
$1 + \dfrac{3}{7}x + \dfrac{{3.6}}{{7.10}}{x^2} + \dfrac{{3.6.9}}{{7.10.13}}{x^3} + \dfrac{{3.6.9.12}}{{7.10.13.16}}{x^4} + ....$

Answer 1

Hint : In the problem given above we need to find the values of $x$ for which the series converge or diverge.. There are various tests that are performed and we will start with d’Alembert’s ratio test which states that the series $S = \sum\limits_{k = 1}^\infty {{x_k}}$ is convergent if there is a $r$ where $r < 1$ for $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = r$ while the series is divergent, if and if $$x = 1$$ then it cannot be concluded whether the series is convergent or divergent. Later on we will apply Raabe’s test.

Complete step by step solution:
Here for the given series
$1 + \dfrac{3}{7}x + \dfrac{{3.6}}{{7.10}}{x^2} + \dfrac{{3.6.9}}{{7.10.13}}{x^3} + \dfrac{{3.6.9.12}}{{7.10.13.16}}{x^4} + ....$
The sum of series will be
${S_n} = 1 + {\rm I}{\rm I}_{i = 1}^\infty \dfrac{{3n{x^n}}}{{(3n + 4)}}$
The ${T_n}$ term of the series will be
${T_n} = \dfrac{{3.6.9.......(3n){x^n}}}{{7.10.13......(3n + 4)}}$
Therefore ${T_{n + 1}}$ term of the series will be
${T_{n + 1}} = \dfrac{{3.6.9.12......(3n)(3n + 3){x^{n + 1}}}}{{7.10.13......(3n + 4)(3n + 7)}}$
Now by using the Ratio test for the given infinite series the equation will be
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{{T_{n + 1}}}}{{{T_n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{(3n + 3)}}{{(3n + 7)}}\dfrac{{{x^{n + 1}}}}{{{x^n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{3 + 3/n}}{{3 + 7/n}}} \right)x = x$
For, $x > 1$ , ${\rm I}{\rm I}_{i = 1}^\infty \dfrac{{3n{x^n}}}{{(3n + 4)}}$ diverges
$x < 1$ , ${\rm I}{\rm I}_{i = 1}^\infty \dfrac{{3n{x^n}}}{{(3n + 4)}}$ converges
Which means that ${S_n} = 1 + {\rm I}{\rm I}_{i = 1}^\infty \dfrac{{3n{x^n}}}{{(3n + 4)}}$ converges for $x < 1$ and ${S_n} = 1 + {\rm I}{\rm I}_{i = 1}^\infty \dfrac{{3n{x^n}}}{{(3n + 4)}}$ diverges for $x > 1$ and for $x = 1$ ratio test fails
Now later on we will apply Raabe’s Test

$$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{{{T_n}}}{{{T_{n + 1}}}} - 1} \right] n = \mathop {\lim }\limits_{n \to \infty } {\text{ n}}\left[ {\dfrac{{3n + 7}}{{3n + 3}} - 1} \right] \\\ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\text{ n}}\left[ {\dfrac{{3n + 7 - 3n + 3}}{{3n + 3}}} \right] = \mathop {\lim }\limits_{n \to \infty } \dfrac{{4n}}{{n(3 + 3/n)}} = \dfrac{4}{3} \\\$$

Since we can see that
$\mathop {\lim }\limits_{n \to \infty } \left[ {\dfrac{{{T_n}}}{{{T_{n + 1}}}} - 1} \right] n = \dfrac{4}{3} > 1$
The series is convergent for $x = 1$

Note : The sum of an infinite sequence of numbers is known as series. The series can be convergent or divergent.
Notes: While solving the above equation always remember that if a number is divided by an infinite value then the results obtained will be equal to zero. Keep in mind that if a ratio test gets failed it does not ensure the nature of the series and apply Raabe’s test also. Here we will also notice that for any positive value of $x$ , the numerator is always greater than the denominator which means each term is greater than $1$ hence the given infinite series is convergent