# Find the sum of the infinite series $\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+…$

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?

• Is $1.2.3.4$ supposed to mean $1 \cdot 2 \cdot 3 \cdot 4$, etc.? – kccu May 10 '17 at 11:01
• It's the product of these numbers – idpd15 May 10 '17 at 11:02
• 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. – John Hughes May 10 '17 at 11:05
• Also: do you know about Taylor series? – John Hughes May 10 '17 at 11:06

$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...=\frac{1}{2}\cdot\frac{1}{1!}+\frac{1}{2^2}\cdot\frac{1}{2!}+\frac{1}{2^3}\cdot\frac{1}{3!}+... = e^\frac{1}{2}-1.$