Re: Convergence/Divergence
A really simple way to do 1 is to determine the ratio of terms. If each term is successively bigger, the sequence is divergent. So take the term for n+1, (2^(2n+2))/(3^(n+3) and divide that by (2^2n)/3^(n+2). You get 4/3. So each term is 4/3 bigger than the last one, so the sequence is divergent.
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