Calc 2
posted by James .
How would I do this one?
n=1 to infinity n^n/4^1+3*n
Does it absolutely converge, conditionally converge, or diverge?
What would I do here?

n^n /[ 4^(1+3n) ] or what?
Need parentheses to see what you mean.
By the way that clearly does not converge. 
This is how it looks: hopefully this clarifies it a bit.
n^n/(4)^1+3*n 
but 4^1 is just 4
do you mean
n^n / (4 + 3n) ?
that would be for large n
n^n /3n
(1/3) n^(n1)
which gets very large indeed
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