Find the sum of the series $\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...$. This type of questions generally require a trick or something and i am not able to figure that out. My guess is that it has something to do with exponential series or binomial series. Any help?

  • $\begingroup$ Is $$ supposed to mean $1 \cdot 2 \cdot 3 \cdot 4$, etc.? $\endgroup$ – kccu May 10 '17 at 11:01
  • $\begingroup$ It's the product of these numbers $\endgroup$ – idpd15 May 10 '17 at 11:02
  • $\begingroup$ Did you try cancelling numerators with denominators? Maybe there's some good reason to have $1,3,5$ in both the numerator and denominator of the third term, but I can't see it. $\endgroup$ – John Hughes May 10 '17 at 11:05
  • $\begingroup$ Also: do you know about Taylor series? $\endgroup$ – John Hughes May 10 '17 at 11:06

Sorry guys, got it.
$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...=\frac{1}{2}\cdot\frac{1}{1!}+\frac{1}{2^2}\cdot\frac{1}{2!}+\frac{1}{2^3}\cdot\frac{1}{3!}+... = e^\frac{1}{2}-1.$
The first equality holds after cancelling the common terms in the numerator and denominator


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