Question

How do you use the ratio test to test the convergence of the series $\sum {\dfrac{{{{11}^n}}}{{(n + 1)\left( {{7^{2n + 1}}} \right)\;}}}$ from $n = 1$ to infinity?

Answer 1

Hint : To test the convergence of the given infinite series, remember the D’Alembert’s ratio test which says that the series $S = \sum\limits_{i = 1}^\infty {{x_i}}$ will be convergence if there exists a $k$ such that $k < 1$ for the $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right| = k$ and the series will be divergent if $k > 1$ for the same limit of the ratio. But this test will fail if we get $k = 1$ that means then we have to go through another test to test the convergence.

Complete step-by-step answer :
In order to test the convergence of the given series we have to find the $nth\;{\text{and}}\;(n + 1)th$ term of the given series that is $\sum {\dfrac{{{{11}^n}}}{{(n + 1)\left( {{7^{2n + 1}}} \right)\;}}}$
We can find $nth\;{\text{and}}\;(n + 1)th$ by just replacing the variable with them separately
So, $nth$ term will be given as
${x_n} = \dfrac{{{{11}^n}}}{{(n + 1)\left( {{7^{2n + 1}}} \right)\;}}$
And $(n + 1)th$ will be given as
${x_{n + 1}} = \dfrac{{{{11}^{n + 1}}}}{{\left( {(n + 1) + 1} \right)\left( {{7^{2(n + 1) + 1}}} \right)\;}} = \dfrac{{{{11}^{n + 1}}}}{{(n + 2)\left( {{7^{2n + 3}}} \right)\;}}$
Now, according to the ratio test, we have to find the following
$k = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x_{n + 1}}}}{{{x_n}}}} \right|$
Putting values, we will get
$k = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\dfrac{{{{11}^{n + 1}}}}{{(n + 2)\left( {{7^{2n + 3}}} \right)\;}}}}{{\dfrac{{{{11}^n}}}{{(n + 1)\left( {{7^{2n + 1}}} \right)\;}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{11(n + 1)}}{{(n + 2)\left( {{7^2}} \right)\;}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{11(n + 1)}}{{(n + 2)49}}} \right|$
Dividing numerator and denominator with $n$ to solve the limit, we will get
$k = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{11\left( {1 + \dfrac{1}{n}} \right)}}{{\left( {1 + \dfrac{2}{n}} \right)49}}} \right|$
Substituting the limits, we will get
$k = \left| {\dfrac{{11(1 + 0)}}{{(1 + 0)49}}} \right| = \dfrac{{11}}{{49}} < 1$
Since $k < 1$ and we know that in the ratio test if the limit comes to be less than one then the series is convergent.
Therefore the series $\sum {\dfrac{{{{11}^n}}}{{(n + 1)\left( {{7^{2n + 1}}} \right)\;}}}$ is convergent.

Note : In this question we have given the general term of the series, so that we put the term numbers which we want and get the terms directly. But in general you will tackle the actual series, so don’t be afraid at that time and firstly find the general term of the series then proceed with the test.