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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.

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  • $\begingroup$ I think you may have meant to write $\cfrac{a_{n+1}}{a_n}\;$ Edit it, as it looks very different. $\endgroup$
    – DonAntonio
    Sep 28, 2016 at 11:06
  • $\begingroup$ @DonAntonio Fixed, thanks. $\endgroup$ Sep 28, 2016 at 13:44
  • $\begingroup$ You're welcome....yet you fixed nothing: it remains the same. $\endgroup$
    – DonAntonio
    Sep 28, 2016 at 17:41

3 Answers 3

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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.

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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:

$$\sqrt[n]{\frac{14^n}{3^{3n+4}(3n+7)}}=\frac{14}{3^3}\cdot\frac1{\sqrt[n]{3^4(3n+7)}}\xrightarrow[n\to\infty]{}\frac{14}{27}\cdot1=\frac{14}{27}<1$$

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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 $n\rightarrow\infty$.

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