Series $\sum_{n=2}^\infty \frac{14^n}{3^{3n+4}(3n+7)}$ Convergence or Divergence Using The Ratio Test I am trying to determine if the following series converges or diverges by using the ratio test, which I believe can be summarized as the following:
$$L=\lim_{n\to \infty}  \left| \frac{a_{n+1}}{a_n} \right|$$
If $L < 1$ then the series converges, if $L > 1$, then it diverges, and if $L = 1$, then it is ambiguous. The series is below:
$$\sum_{n=2}^\infty \frac{14^n}{3^{3n+4}(3n+7)}$$
I understand the ratio test in theory, but am not sure how to put it into practice for a series like this.
 A: Hint:
$$\sum_{n=2}^\infty \frac{14^n}{3^{3n+4}(3n+7)}\leq\sum_{n=2}^\infty \frac{14^n}{7\cdot 3^{3n+4}}=\frac{1}{7}\sum_{n=2}^\infty \frac{14^n}{3^4\cdot 27^n}=\frac{1}{7\cdot 81}\sum_{n=2}^\infty \left(\frac{14}{27}\right)^n$$
Convergence follows from dominant convergence of the infinite gemetric series and the fact that all involved terms are positive.
You can also directly apply the ratio test:
$$a_{n+1}/a_{n}=\frac{14^{n+1}}{3^{3n+7}(3n+10)}\cdot\frac{3^{3n+4}(3n+7)}{14^{n}}=\frac{14}{27}\frac{3n+7}{3n+10}\leq14/27=q<1$$
The last inequality comes from the fact that $3n+7\leq 3n+10$. As we found a value for $q<1$, we know that the series converges by the ratio rule.
A: Ratio test:
$$\frac{14^{n+1}}{3^{3n+7}(3n+10)}\cdot\frac{3^{3n+4}(3n+7)}{14^n}=\frac{14}{3^3}\cdot\frac{3n+7}{3n+10}\xrightarrow[n\to\infty]{}\frac{14}{27}\cdot1=\frac{14}{27}<1$$
$\;n\,-$ th root test:
$$\sqrt[n]{\frac{14^n}{3^{3n+4}(3n+7)}}=\frac{14}{3^3}\cdot\frac1{\sqrt[n]{3^4(3n+7)}}\xrightarrow[n\to\infty]{}\frac{14}{27}\cdot1=\frac{14}{27}<1$$
A: Here we have
$$a_n=\dfrac{14^n}{3^{3n+4}(3n+7)}$$
so that
$$a_{n+1}=\dfrac{14^{n+1}}{3^{3(n+1)+4}(3(n+1)+7)}=\dfrac{14\cdot 14^n}{3^{3n+4}\cdot 3^3(3n+10)}$$
Try to compute the limit of $a_{n+1}/a_n$ as $n\rightarrow\infty$.
