# Thread: Ratio Test for Convergence

1. ## Ratio Test for Convergence

Having a bit of trouble with this one. Can anyone help?
Many thanks.

Q. $\displaystyle \sum_{n=1}^{\infty}\frac{(n-1)!}{2^n}$

Attempt: $\displaystyle U_n = \frac{(n-1)!}{2^n} => u_n_+_1 = \frac{n!}{2^n^+^1}$
$\displaystyle \lim_{n \to \infty} |\frac{U_n_+_1}{U_n}| => \lim_{n \to \infty} |\frac{n!}{2^n^+^1}.\frac{2^n}{(n-1)!}| => \lim_{n \to \infty} |\frac{n!}{2^n.2}.\frac{2^n}{(n-1)n!}| => \lim_{n \to \infty}|\frac{1}{2(n-1)}| => \lim_{n \to \infty}|\frac{1}{2n-2}|$

Ans.: (From text book): Series is divergent.

2. ## Re: Ratio Test for Convergence

Originally Posted by GrigOrig99
Having a bit of trouble with this one. Can anyone help?
Many thanks.

Q. $\displaystyle \sum_{n=1}^{\infty}\frac{(n-1)!}{2^n}$

Attempt: $\displaystyle U_n = \frac{(n-1)!}{2^n} => u_n_+_1 = \frac{n!}{2^n^+^1}$
$\displaystyle \lim_{n \to \infty} |\frac{U_n_+_1}{U_n}| => \lim_{n \to \infty} |\frac{n!}{2^n^+^1}.\frac{2^n}{(n-1)!}| => \lim_{n \to \infty} |\frac{n!}{2^n.2}.\frac{2^n}{(n-1)n!}| => \lim_{n \to \infty}|\frac{1}{2(n-1)}| => \lim_{n \to \infty}|\frac{1}{2n-2}|$

Ans.: (From text book): Series is divergent.
$\displaystyle \left|\frac{n!}{2^n^+^1}.\frac{2^n}{(n-1)!}\right|=\frac{n}{2}$

3. ## Re: Ratio Test for Convergence

Great. Thank you.