# Yonglie's question via email about ratio test

#### Prove It

##### Well-known member
MHB Math Helper
Use the ratio test to test the convergence of the positive term series \displaystyle \begin{align*} \sum_{n = 1}^{\infty}{\frac{2n+3}{4n^3 + n}} \end{align*}
First, because the series is positive term, we don't have to worry about absolute values. Now \displaystyle \begin{align*} a_n = \frac{2n + 3}{4n^3 + n} \end{align*} and

\displaystyle \begin{align*} a_{n + 1} &= \frac{2\left( n + 1 \right) + 3}{4 \left( n + 1 \right) ^3 + n + 1} \\ &= \frac{2n + 2 + 3}{4 \left( n^3 + 3n^2 + 3n + 1 \right) + n + 1} \\ &= \frac{ 2n + 5}{4n^3 + 12n^2 + 12n + 4 + n + 1} \\ &= \frac{2n + 5}{4n^3 + 12n^2 + 13n + 5} \end{align*}

so the ratio is

\displaystyle \begin{align*} \frac{a_{n + 1}}{a_n} &= \left( \frac{2n + 5}{4n^3 + 12n^2 + 13n + 5} \right) \left( \frac{4n^3 + n}{2n + 3} \right) \\ &= \frac{8n^4 + 2n^2 + 20n^3 + 5n}{8n^4 + 12n^3 + 24n^3 + 36n^2 + 26n^2 + 39n + 10n + 15} \\ &= \frac{8n^4 + 20n^3 + 2n^2 + 5n}{8n^4 + 36n^3 + 62n^2 + 49n + 15} \end{align*}

and the limit of the ratio is

\displaystyle \begin{align*} \lim_{n \to \infty} \frac{a_{n + 1}}{a_n} &= \lim_{n \to \infty} \frac{8n^4 + 20n^3 + 2n^2 + 5n}{8n^4 + 36n^3 + 62n^2 + 49n + 15} \\ &= \lim_{n \to \infty} \frac{n^4 \left( 8 + \frac{20}{n} + \frac{2}{n^2} + \frac{5}{n^3} \right) }{n^4 \left( 8 + \frac{36}{n} + \frac{62}{n^2} + \frac{49}{n^3} + \frac{15}{n^4} \right) } \\ &= \lim_{n \to \infty} \frac{8 + \frac{20}{n} + \frac{2}{n^2} + \frac{5}{n^3}}{8 + \frac{36}{n} + \frac{62}{n^2} + \frac{49}{n^3} + \frac{15}{n^4} } \\ &= \frac{8 + 0 + 0 + 0}{8 + 0 + 0+0+0} \\ &= \frac{8}{8} \\ &= 1 \end{align*}

Since the limit of the ratio is 1, the ratio test is INCONCLUSIVE and we are unable to make a statement about the convergence or divergence of the series.

But a simple comparison:

\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{2n + 3}{4n^3 + n} &= \sum_{n = 1}^{\infty} \frac{n \left( 2 + \frac{3}{n} \right) }{ n \left( 4n^2+ 1 \right) } \\ &= \sum_{n = 1}^{\infty} \frac{2 + \frac{3}{n}}{4n^2 + 1} \end{align*}

has the same behaviour as \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{2}{4n^2} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{n^2} \end{align*} which is a convergent p-series.

So by limit comparison, \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{2n + 3}{4n^3 + n} \end{align*} is convergent.