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Kindly asking, what can I do about series

$$ \left(\frac{1}{3}\right)^2+\left(\frac{1\times 4}{3\times 6}\right)^2+\left(\frac{1\times 4\times 7}{3\times 6\times 9}\right)^2+...+\left(\frac{1\times 4\times 7\times...\times (3n-2)}{3\times 6\times 9\times...\times3n}\right)^2+...$$

Indeed, the ratio test fails. Thank you.

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Not sure if this fact is useful, but the fractions are the terms in the expansion of $$(1-x)^{-\frac{1}{3}}$$. – picakhu Nov 27 '12 at 22:00
up vote 3 down vote accepted

Let $a_n$ be the $n$th term. The ratio of successive terms is $$\frac{a_{n+1}}{a_n} = \frac{(3n+1)^2}{9(n+1)^2} = 1 - \frac{4}{3n} + O\left(\frac{1}{n^2}\right).$$ Thus, the series converges by Raabe's test, which tells us that if $$\left|\frac{a_{n+1}}{a_{n}}\right| \sim 1 - \frac{s}{n} \hspace{5ex}(n\to\infty),$$ then the series converges absolutely if $s>1$, diverges if $s<1$, and may converge or diverge if $s=1$.

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$$\frac{1\times 4\times 7\times\cdots\times (3n-2)}{3\times 6\times 9\times\cdots\times3n}\leqslant\frac1{(n+1)^{2/3}}$$

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Could you elaborate? – Pedro Tamaroff Nov 27 '12 at 23:18
@PeterTamaroff Sure: one proves this inequality the obvious way (induction on $n$) and one uses it the obvious way (square it and compare the series in the post to the Riemann series of exponent $4/3$). – Did Nov 28 '12 at 6:25

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