Product of terms in a series

Question

Okay so this question requires me to sum this series. My problem is however the product of the odd terms that appear in the sequence. I'm not able to write a general term for this sequence.

$$1+\frac{1}{1!}\cdot\frac{1}{4}+\frac{1\cdot3}{2!}\cdot(\frac{1}{4})^2+\frac{1\cdot3\cdot5}{3!}\cdot(\frac{1}{4})^3+........... \infty$$

• The $\cdot$ represents the product of the two terms.

• $!$ represents the factorial of the specified number.

My doubt is that after having the general term, how do I manipulate it to get the final sum? It makes no sense to me still.

Any help would be appreciated.Thanks for giving this your time.

Answer 1

Hint:

$$1 \times 3 \times 5 = \frac{6!}{2(4)(6)}= \frac{(2 \times 3)!}{2^3(3!)}$$

Answer 2

$$a_n = 1\cdot3\cdot\dots\cdot(2n-1)\cdot\frac{1}{n!}\frac1{4^n} = \frac{1\cdot2\cdot3\cdot4\dots\cdot2n}{2\cdot4\cdot\dots2n}\frac{1}{n!}\frac1{4^n}=\frac{(2n)!}{2^n(n!)}\frac{1}{n!}\frac1{4^n}$$ $$=\frac{(2n)!}{n!n!}\left(\frac18\right)^n={2n\choose n}\left(\frac18\right)^n$$ So we have to evaluate $$\sum_{n=0}^{\infty}{2n\choose n}\left(\frac18\right)^n$$ Now using $$(1-4x)^{-\frac12}=\sum_{n=0}^{\infty}(-4)^n{-\frac12\choose n}x^n=\sum_{n=0}^{\infty}{2n\choose n}x^n$$ We can conclude $$\sum_{n=0}^{\infty}{2n\choose n}\left(\frac18\right)^n=\sqrt2$$