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 March 19th, 2019, 06:37 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Series sum The sum of series $$\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^3}+\frac{5\cdot 7\cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty$$
 March 19th, 2019, 10:04 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,550 Thanks: 1401 should the denominator of the 2nd term be $3! \cdot 3^2$ ? Otherwise I don't see the pattern.
March 19th, 2019, 10:43 AM   #3
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Thanks Romsek

Quote:
 Originally Posted by panky The sum of series $$\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7\cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty$$

 March 27th, 2019, 01:55 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 The sum can be written as $\displaystyle \sum_{n=2}^{\infty} \frac{\prod_{n=2}^{\infty} (1+2n) }{3^{n-1} n! }$ . To compute it seems complicated.
 March 27th, 2019, 02:51 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 The sum should be written as $\displaystyle \sum_{n=2}^\infty \frac{\displaystyle \prod_{k=2}^n (2k + 1)}{3^{n-1}n!}$, which evaluates to 3√3 - 2 according to W|A. Thanks from topsquark and idontknow

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