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March 19th, 2019, 06:37 AM   #1
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Series sum

The sum of series $$\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^3}+\frac{5\cdot 7\cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty$$
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March 19th, 2019, 10:04 AM   #2
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should the denominator of the 2nd term be $3! \cdot 3^2$ ?

Otherwise I don't see the pattern.
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March 19th, 2019, 10:43 AM   #3
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Thanks Romsek

Quote:
Originally Posted by panky View Post
The sum of series $$\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7\cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty$$
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March 27th, 2019, 01:55 PM   #4
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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.
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March 27th, 2019, 02:51 PM   #5
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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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