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I have to find the radius of convergence and see what happens at both ends of the interval of convergence for the following power series:

$$ \sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n+1)!}(x-3)^n $$

I found the radius of convergence to be $\mid x - 3\mid \lt \frac{4}{27}$ by applying the ratio test. I have to find out whether the series is convergent (absolutely, uniformly) or not when $x=\frac{77}{27}$ and when $x=\frac{85}{27}$ (the ends of the interval of convergence).

When $x=\frac{77}{27}$, I get an alternating series:

$$ \sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n+1)!}\bigg(-\frac{4}{27}\bigg)^n $$

So, I have to apply the Leibnitz test, but I have problems finding: $$ \lim_{n\to\infty}\frac{(3n)!}{n!(2n+1)!}\bigg(\frac{4}{27}\bigg)^n=? $$

because I get: $$ \frac{\infty}{\infty}0 $$

And I can't show that the series decreases monotonically. I have similar problems when $x=\frac{85}{27}$. I'd appreciate the help.

When $x=\frac{85}{27}$, I tried applying the ratio test and got $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, so the test tells me nothing, so I applied the Raabe's test and got

$$ \lim_{n\to\infty}\bigg[n\bigg(\frac{a_n}{a_{n+1}}-1\bigg)\bigg]=\frac{3}{2}\gt1 $$

which means that the series is convergent, so I'm still struggling with $x=\frac{77}{27}$

Daniel
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  • For $$ \sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n+1)!}(\frac{-4}{27})^n $$ Could you use the ratio test? The Leibnitz test yields a large mass of calculations. –  Mar 23 '14 at 15:53
  • The ratio test works only for series with non-negative terms, here I have an alternating series, I have to apply the Leibnitz test. – Daniel Mar 23 '14 at 16:04
  • My bad. Sorry.${}$ –  Mar 23 '14 at 16:05
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    @Daniel Since for $x = \frac{85}{27}$, the terms are the absolute value of the corresponding term on the other end, if the series converges there, it converges absolutely on the other end. – Daniel Fischer Mar 23 '14 at 16:33
  • @DanielFischer Game over. :-) – Did Mar 23 '14 at 16:36
  • So the series is convergent for $x\in\bigg(\frac{77}{27};\frac{85}{27}\bigg]$ and absolutely convergent for $x=\frac{77}{27}$, right? – Daniel Mar 23 '14 at 16:41

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