Question

Find whether the following series is convergent or divergent.
$1 + 3x + 5{x^2} + 7{x^3} + 9{x^4} + ...$

Answer 1

We can check the convergence of this series using a ratio test. For that we have to identify the general term of the series. Then according to the conclusion of the test we can find the series is convergent or divergent.

Formula used: Let $\sum\limits_{i = 1}^n {{a_n}}$ be a series. Suppose there exists $r$ such that $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{a_{n - 1}}}} = r$.
If $r < 1$, then the series is convergent.
If $r > 1$, then the series is divergent.
If $r = 1$, the series may converge or diverge.

Complete step-by-step answer:
The series given is $\sum\limits_{n = 1}^\infty {{a_n}} = 1 + 3x + 5{x^2} + 7{x^3} + 9{x^4} + ...$
We have the terms ${a_1} = 1,{a_2} = 3x,{a_3} = 5{x^2},...$
So, in general the ${n^{th}}$ term ${a_n} = (2n - 1){x^{n - 1}}$
Similarly, the ${(n - 1)^{th}}$ term ${a_{n - 1}} = [2(n - 1) - 1]{x^{n - 1 - 1}} = (2n - 3){x^{n - 2}}$
So, we can calculate the ratio $\dfrac{{{a_n}}}{{{a_{n - 1}}}}$.
$\dfrac{{{a_n}}}{{{a_{n - 1}}}} = \dfrac{{(2n - 1){x^{n - 1}}}}{{(2n - 3){x^{n - 2}}}}$
We have $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$.
$\Rightarrow \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \dfrac{{(2n - 1)}}{{(2n - 3)}}{x^{n - 1 - (n - 2)}}$
$\Rightarrow \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \dfrac{{2n - 1}}{{2n - 3}}{x^{}}$
Applying limits,
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{2n - 1}}{{2n - 3}}x$
Dividing Numerator and Denominator by $n$ we have,
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{2 - \dfrac{1}{n}}}{{2 - \dfrac{3}{n}}}x$
As $n \to \infty ,\dfrac{1}{n} \to 0$
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \dfrac{{2 - 0}}{{2 - 0}}x = x$
Hence by ratio test conclusion, the sequence is convergent if $x < 1$ and divergent if $x > 1$.
We can check the case of $x = 1$ separately.
Let $x = 1$.
Then the series becomes $1 + 3 + 5 + 7 + 9 + ...$ , the odd number series and is clearly divergent.
So, the given series is convergent for all values of $x$ less than $1$ and divergent for all values of $x$ greater than or equal to $1$.

So, the correct answer is “Option A”.

Additional Information: There are other types of test for convergence namely Root test, Comparison test, Dirichlet’s test etc. There is also another concept named absolute convergence related to convergence.

Note: A series can be tested convergent or divergent by different means. It is important to identify the appropriate option in each problem and check its conditions. If there are any cases which conclusion is not applicable that case must be checked separately.