Showing the divergence of the series where $a_1 = 2$ and $a_{n+1} = \frac{5n+1}{4n+3}a_n$.

Question

Consider a series such that its $i$th term $a_i$ is defined by $a_1 = 2$ and $a_{n+1} = \dfrac{5n+1}{4n+3}a_n$.

I would like to show that this series is divergent. Here's how I thought about it: notice $$\begin{align} &a_2 = \left(\dfrac{11}{11}\right)(2) = 2\\ &a_3 = \left(\dfrac{16}{15}\right)(2) \\ &a_4 = \left(\dfrac{21}{19}\right)\left(\dfrac{16}{15}\right)(2) \\ &\vdots\\ &a_k = \dfrac{16 \cdot 21 \cdots (5k+1)}{15 \cdot 19 \cdots (4n+3)}(2)\to \dfrac{5}{4}\cdot (\text{stuff }> 0) \neq 0 \end{align}$$ Hence the series is divergent.

I'm not sure if my approach above is correct. Is there a more rigorous way to do this under a time crunch (think Math GRE Subject Test)?

Answer 1

By the ratio test, since a(n+1)/a(n) = (5n+1)/(4n+3), this expression tends towards 5/4 as n tends towards infinity. Since the value of 5/4 greater than 1, your sequence diverges.

Answer 2

$a_n>0$ and $a_{n+1}=\frac{5n+1}{4n+3}a_n>\frac{9}{8}a_n$ when $n\geq5$, so $a_n>(\frac{9}{8})^{n-5}a_5\rightarrow+\infty$ as $n\rightarrow\infty$.

Answer 3

Hints:

$\sum_{n=1}^{\infty}a_n$ exists then $\lim_{n\to \infty}a_n=0$.