Monday 27 November 2017

calculus - Problem solving a series with convergence test: $sum_{k=1}^{infty}frac{k!}{(2k)!}$



Good morning, I have a big problem solving this:
$\sum_{k=1}^{\infty}\frac{k!}{(2k)!}\:$



I'm trying solving this limit with test of D'Alembert, but I have a problem solving the limit.



$\lim_{k\rightarrow\infty}\frac{(k+1)!k!}{(2k+2)!2k!}=(?)$




please, help me.


Answer



We have that
$$
\lim_{k\to\infty}\frac{(k+1)!(2k)!}{k!(2(k+1))!}=\lim_{k\to\infty}\frac{(k+1)(2k)!}{(2k+2)!}=\lim_{k\to\infty}\frac{k+1}{(2k+2)(2k+1)}=\lim_{k\to\infty}\frac1{2(2k+1)}=0.
$$
Hence, the series converges by the ratio test.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...