$$ 1+\frac{4}{6}+\frac{4\cdot5}{6\cdot9}+\frac{4\cdot5\cdot6}{6\cdot9\cdot12}+\cdots $$
I could reduce it to $n$th term being $\dfrac{(n+1)\cdot(n+2)}{n!\cdot3^n}$. Took me an hour just to get to this. But I am now stuck up. PL. Help
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1See http://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot – lab bhattacharjee Dec 19 '14 at 17:25
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The $n$th term $$=\frac{4\cdot5\cdot6\cdots(n+3)}{6\cdot9\cdot12\cdots(3n+3)},$$ right? – lab bhattacharjee Dec 19 '14 at 17:35
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Have corrected nth term in the question. – Rohan Dec 19 '14 at 17:49
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It's incorrect, we don't need it particularly though if you follow the linked answers. – lab bhattacharjee Dec 19 '14 at 17:53
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Any other method ? – Rohan Dec 19 '14 at 17:59
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$$1+\frac{4}{6}+\frac{4\cdot5}{6\cdot9}+\frac{4\cdot5\cdot6}{6\cdot9\cdot12}+\cdots$$ $$=1+\sum\limits_{n=1}^\infty\frac{(n+3)!}{2\times3\times3^n\times(n+1)!}$$ $$=1+\frac{1}{2}\sum\limits_{n=1}^\infty\frac{(n+2)\times(n+3)}{3^{n+1}}$$
By induction, we can show that for $n\ge7$, $0<\frac{(n+2)\times(n+3)}{3^{n+1}}<\frac{1}{n^2}$, and hence the series is convergent.
In fact, $=1+\frac{1}{2}\cdot\frac{11}{4}=\frac{19}{8}$
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