Find Sum of Series $$2+\frac5{2!\cdot3}+\frac{5\cdot7}{3!\cdot3^2}+\frac{5\cdot7\cdot9}{4!\cdot3^3}+ \cdots\cdots $$
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?
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 $$ \sum_{n=1}^\infty a_n = 3\sqrt{3}. $$