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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The ratio test should work...
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?
You need to evaluate $\displaystyle \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$.
oh shucks i totally forgot about that. was too concerned with the ratios. thanks!!!
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