I'm trying to determine whether the sum $$S=\frac{2}{1}+\frac{2\cdot 5}{1\cdot 5}+\frac{2\cdot 5\cdot 8}{1\cdot 5\cdot 9}+...+\frac{2\cdot 5\cdot 8...(3n-1)}{1\cdot 5\cdot 9...(4n-3)}+...$$ converges or not.

Even if it does converge, I am not interested in what it is equal to; I just need to show it converges or diverges.

I'm drawing a blank here, but my gut instinct tells me "no." I don't think that $a_n$ (the $n^{th}$ term in the sum) approaches $0$ when $n$ approaches infinity, which is something that must be met for all converging sums, but I don't know how to prove it, as this is a rather difficult limit.

Does this sum converge or not? and how would you go about showing your answer?

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    $\begingroup$ Try the ratio test? $\endgroup$ – Daniel Fischer Oct 31 '14 at 21:19
  • $\begingroup$ I have not heard of that, we didn't get that far in class, just googled it now and it does indeed solve the question easily. thank you Daniel. $\endgroup$ – Oria Gruber Oct 31 '14 at 21:20
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    $\begingroup$ For a better intuitive feel, note that for large $n$ $\dfrac{3n-1}{4n-3}$ is approximately $\dfrac34$, so that eventually each term is approximately three-quarters of the previous one. $\endgroup$ – Brian M. Scott Oct 31 '14 at 21:21
  • $\begingroup$ This is a binomial series. $\endgroup$ – Lucian Oct 31 '14 at 21:58

Let $a_n = \frac{2.5.8...(3n-1)}{1.5.9...(4n-3)}$. And notice that

$$\begin{align}\lim_{n\to\infty}\Big|\frac{a_{n+1}}{a_n}\Big| &= \lim_{n \to \infty}\frac{2.5.8...(3n-1)(3(n+1) - 1)}{1.5.9...(4(n+1)-3)}\frac{1.5.9...(4n-3)}{2.5.8...(3n-1)} \\&= \lim_{n\to\infty}\frac{3n -2}{4n-1} = \frac{3}{4} < 1 \end{align}$$

Use D'Alembert Test.


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