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 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  
March 27th, 2019, 01:55 PM  #4 
Senior Member Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 
The sum can be written as $\displaystyle \sum_{n=2}^{\infty} \frac{\prod_{n=2}^{\infty} (1+2n) }{3^{n1} n! }$ . 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 $\displaystyle \sum_{n=2}^\infty \frac{\displaystyle \prod_{k=2}^n (2k + 1)}{3^{n1}n!}$, which evaluates to 3√3  2 according to WA.


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