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$$\sum_{n=1}^{\infty}\frac{1\cdot 4\cdot 7\cdots (3n+1)}{n^5}$$

this question comes right after the question that asks me to prove it with the limit comparsion test. I need to prove that it's divergent.

The limit comparsion test states that in $\sum a_n$, $\sum b_n$, if $\lim \frac{a_n}{b_n}=0$, then $\sum b_n$ convergent $\implies$$\sum a_n$ convergent. How can I modify this theorem to prove divergence? Also, is is possible to prove the series above is divergent using this test?

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  • $\begingroup$ It seems offhand the term being summed doesn't approach zero. [factorials increase faster than powers] $\endgroup$ – coffeemath May 28 '16 at 3:13
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I like to write expressions with "..." in terms os $\sum$ and $\prod$.

In this case

$\begin{array}\\ \dfrac{1\cdot 4\cdot 7\cdots (3n+1)}{n^5} &=\dfrac{\prod_{k=0}^n (3k+1)}{n^5}\\ &\gt\dfrac{\prod_{k=1}^n (3k+1)}{n^5}\\ &\gt\dfrac{\prod_{k=1}^n (3k)}{n^5}\\ &=\dfrac{3^nn!}{n^5}\\ \end{array} $

and this diverges extremely rapidly - so much so that it will diverge for any $n^m$ for fixed $m$ in the denominator.

As a matter of fact, since $3 > e$ and $n! > (n/e)^n$, this still diverges if the denominator is $n^n$.

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  • $\begingroup$ isn't there a way to do it that requires the divergence of a more simple series? The following exercises in my list ask me to prove the divergence of one similar to this you showed me. $\endgroup$ – Guerlando OCs May 28 '16 at 3:34
  • $\begingroup$ or you're talking about divergence of the limit of $\dfrac{3^nn!}{n^5}$? in this case I think I can use it. $\endgroup$ – Guerlando OCs May 28 '16 at 3:35
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The ratio test zaps this one right off the bat. $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{(2n+4)}{(n+1)^5}\frac{n^5}1=\lim_{n\rightarrow\infty}\frac{2n+4}{\left(1+\frac1n\right)^5}=\infty$$ So the series diverges. Note how the ratio test created a context where most of the factors of the product canceled out.

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