how do you prove using ratio test that the series from n=1 to infinity of

((n+1)^2 )(2^n) / n! converges?

i tried using a_n+1 over a_n but cant seem to get the answer

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- Apr 11th 2011, 05:05 AMalexandrabel90ratio
how do you prove using ratio test that the series from n=1 to infinity of

((n+1)^2 )(2^n) / n! converges?

i tried using a_n+1 over a_n but cant seem to get the answer - Apr 11th 2011, 05:22 AMProve It
The ratio test should work...

- Apr 11th 2011, 05:49 AMalexandrabel90
this is what i got:

by ratio test,

l a_n+1 / a_n l = ((n+2)^2 )(2^n+1) / ((n+1)! ) / ((n+1)^2 )(2^n) / n!

= [1 + 1/ (n+1) ]^2 multiply with (2) divide by (n+1)

how can i tell that this is less than 1? - Apr 11th 2011, 05:53 AMProve It
You need to evaluate $\displaystyle \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$.

- Apr 11th 2011, 05:56 AMalexandrabel90
oh shucks i totally forgot about that. was too concerned with the ratios. thanks!!!