Question

How do you apply the ratio test to determine if $\sum{\dfrac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2n-1 \right)}{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3n-2 \right)}}$ from $n=\left[ 1,\infty \right)$ is convergent to divergent? $$$$

Answer 1

We recall infinite series , convergence and divergence of an infinite series. We use d'Alembert's ratio test which states that the series $S=\sum\limits_{k=1}^{\infty }{{{x}_{k}}}$ is convergent if there exists an $r$ such that $r<1$ for $\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}_{n+1}}}{{{x}_{n}}} \right|=r$. If $r>1$ the series is divergent and if $r=1$ we cannot conclude whether the series is divergent or convergent.

Complete step-by-step solution:
We know that the sum of terms in an infinite sequence is called an infinite series which is given by
$S={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...=\sum\limits_{k=1}^{\infty }{{{x}_{k}}}$
The ${{n}^{\text{th}}}$ partial sum of infinite series is the sum of first $n$ terms that is ${{S}_{n}}=\sum\limits_{k=1}^{n}{{{x}_{k}}}$. An infinite series is said to be convergent when its partial sums ${{S}_{1}},{{S}_{2}},{{S}_{3}},...$ tends to a limit. Mathematically if $l$ is the limit and for arbitrary positive small number $\varepsilon$there exits ${{n}_{0}}\in \mathsf{\mathbb{N}}$ such that
$\left| {{S}_{n}}-{{n}_{0}} \right|<\varepsilon$
If such a limit $l$ does not exist we call the series divergent. We are given the following series in the question
$\sum{\dfrac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2n-1 \right)}{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3n-2 \right)}}$
Let us apply d'Alembert's ratio test for the above series. Let us consider $r$ whether

& r=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| \\\ & \Rightarrow r=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2\left( n+1 \right)-1 \right)}{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3\left( n+1 \right)-2 \right)}}{\dfrac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2n-1 \right)}{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3n-2 \right)}} \right| \\\ & \Rightarrow r=\displaystyle \lim_{n \to \infty }\left| \dfrac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2\left( n+1 \right)-1 \right)}{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3\left( n+1 \right)-2 \right)}\times \dfrac{1\cdot 4\cdot 7\cdot \cdot \cdot \left( 3n-2 \right)}{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2n-1 \right)} \right| \\\ \end{aligned}$$ We cancel out like terms from the numerators and denominators to have $$\begin{aligned} & \Rightarrow r=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( 2\left( n+1 \right)-1 \right)}{\left( 3\left( n+1 \right)-2 \right)} \right| \\\ & \Rightarrow r=\displaystyle \lim_{n \to \infty }\left| \dfrac{2n+1}{3n+1} \right| \\\ \end{aligned}$$ We divide both numerator and denominator of the limit by $n$ to have; $$\begin{aligned} & \Rightarrow r=\displaystyle \lim_{n \to \infty }\left| \dfrac{2+\dfrac{1}{n}}{3+\dfrac{1}{n}} \right| \\\ & \Rightarrow r=\left| \dfrac{2}{3} \right|=\dfrac{2}{3}<1 \\\ \end{aligned}$$ So the series is convergent. $$$$ **Note:** We note that since we are given $n=\left[ 1,\infty \right)$ we do not need to consider the left hand limit on $n$. There are different types of roots test too like Cauchy’s root test $r=\displaystyle \lim_{n \to \infty }{\mathop{\lim \sup }}\,\sqrt[n]{\left| {{a}_{n}} \right|}$ where $r$ has to be less than $1$ for convergence and limit comparison test which states that if $\left\\{ {{a}_{n}} \right\\},\left\\{ {{b}_{n}} \right\\}>0$ and $\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}$ exist ,is finite and non-zero then $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ diverges if $\sum\limits_{n=1}^{\infty }{{{b}_{n}}}$ diverges. A sequence $\left\\{ {{x}_{n}} \right\\}$ is convergent if there exist a limit $l$ such that $\displaystyle \lim_{n \to \infty }{{x}_{n}}\to l$. If the sequence is convergent it does not imply the convergence of series for example $\left\\{ \dfrac{1}{n} \right\\}$ converges to 0 but the series $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}$ is divergent.