I have a series $$\sum_{n=1}^\infty\left(\frac{2^n n!}{5\cdot7\cdot9\cdots(2n+3)}\right)^p$$ and I want to know in what value of $p$, this series converges.

So I applied the ratio test, but the limit was 1, which does not provide any information.

And then, I modified this into the equivalent series: $$\sum_{n=1}^\infty\left(\frac{2\cdot4\cdot6\cdots(2n)}{5\cdot7\cdot9\cdots(2n+3)}\right)^p$$ But I don't know what I should do now.

Not restricted in this problem, although I know many convergence tests, in many problems such this, usually I don't know how to determine the convergence of the series.

Please give me an advice.
Thank you.

  • 1
    $\begingroup$ Where does the problem come from? $\endgroup$ – Gerry Myerson Jun 11 '13 at 12:18
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    $\begingroup$ $$\frac{n!}{(n+\frac{3}{2})!} \sim \frac{1}{n^{3/2}}, $$ so the series converges if and only if $p > \frac{2}{3}$. $\endgroup$ – Sangchul Lee Jun 11 '13 at 12:26
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    $\begingroup$ You might want to try the Raabe–Duhamel test. It is listed among convergence tests on Wikipedia, but doesn't have an entry of its own on WP. $\endgroup$ – Harald Hanche-Olsen Jun 11 '13 at 12:28
  • $\begingroup$ @Harald Hanche-Olsen : I checked Raabe-Duhamel test works for this! Thank you. $\endgroup$ – Analysis Jun 11 '13 at 12:48
  • $\begingroup$ And are there some tips for the convergence test? It's quite hard for me to find the method for solving the problem... I think I don't have an intuition. :( $\endgroup$ – Analysis Jun 11 '13 at 12:50

Do the following.

$${2^n n! \over 5\cdot 7 \cdot 9\cdots (2n + 3) } = {3\cdot 4^n n!n!\over (2n)! (2n + 1)(2n + 3) }.$$ Now try using Stirling's Formula to analyze the summands with the root test.


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