Convergence of a series

Nov 2009
23
0
Hi,
How can i check if this series convergent or not?

(3n choose n)(1/3)^n, and the sum is running from n=0 to infinity.

Thanks for any help...
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Hi,
How can i check if this series convergent or not?

(3n choose n)(1/3)^n, and the sum is running from n=0 to infinity.

Thanks for any help...
use the ratio test ... remember that \(\displaystyle \displaystyle \left( \begin{array}{c} x \\ y \end{array} \right) = \frac{x!}{y!(x-y)!}\)
 
Dec 2009
1,506
434
Russia
d'Alembert limit test.

\(\displaystyle \Sigma^\infty_{n=1}{ {3n}\choose{n}}(\frac{1}{3})^n\)

By d'Alembert limit test, if \(\displaystyle lim_{n\to\infty} \frac{a_{n+1}}{a_n}=L\)
and L<1 then your series is converges, if L>1 then your series is divergence and if L=1 then the test is not helping you.

\(\displaystyle lim_{n\to\infty} \frac{a_{n+1}}{a_n}=lim_{n\to\infty} \frac{(3n+3)!(2n)!n!{(\frac{1}{3}})^{n+1}}{(2n+2)!(n+1)!(3n)!{(\frac{1}{3}})^{n}}=\frac{1}{3}lim_{n\to\infty} \frac{(3n+1)(3n+2)(3n+3)}{(2n+1)(2n+2)(n+1)}=\frac{1}{3}\frac{27}{4}=\frac{9}{4}>1=L\)
Hence, your series is diverges!