Summation of series with factorial

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I tried breaking the terms into differences or finding a generalised term but did not get it right. Can someone please help me to proceed with this?

• Are you sure the first term is $2$, and not $1$? – Clement C. Nov 4 '17 at 22:08

Assuming the first term is $1$ (and not $2$ as written), the general term of the series is, for $n\geq 1$, $$a_n \stackrel{\rm def}{=} \frac{\prod_{k=2}^{n}(2k+1)}{n!3^{n-1}} = \frac{\prod_{k=1}^{n}(2k+1)}{n!3^{n}} = \frac{(2n+1)!}{n!3^{n}\prod_{k=1}^n(2k)} = \frac{(2n+1)!}{n!3^{n}2^nn!} = \frac{(2n+1)!}{(n!)^26^{n}}$$ or, equivalently, $a_n= \binom{2n}{n}\left(\frac{1}{6}\right)^n$.
Now, either you work towards finding the general form for $$f(x) = \sum_{n=1}^\infty (2n+1) \binom{2n}{n}x^n$$ (a power series with radius of convergence $1/4$), which you can find by relating it to both $g(x) = \sum_{n=1}^\infty n\binom{2n}{n}x^{n-1}$ (recognize a derivative) and $h(x) = \sum_{n=1}^\infty \binom{2n}{n}x^{n}$, since $$f(x) = 2xg(x)+h(x)\,;$$ or, by other means (there may be?) you establish that $f(1/6) = 3\sqrt{3}$, leading to $$\sum_{n=1}^\infty a_n = 3\sqrt{3}.$$
• If $n=1$, the product is empty, not 5 since it goes from 2 to $n$. – marty cohen Nov 4 '17 at 22:43
• @martycohen for $n=1$, you get 1, the purported first term. $n=2$ gives 5/(6.3) – Clement C. Nov 4 '17 at 22:45