Ratio Test for Convergence

**GrigOrig99** · April 26th 2012, 10:15 AM

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

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

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

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

**Plato** · April 26th 2012, 11:09 AM

Originally Posted by GrigOrig99

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

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

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

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

$\left|\frac{n!}{2^n^+^1}.\frac{2^n}{(n-1)!}\right|=\frac{n}{2}$

**GrigOrig99** · April 26th 2012, 11:33 AM

Great. Thank you.

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