# Calc 2

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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?

• Calc 2 -

n^n /[ 4^(1+3n) ] or what?
Need parentheses to see what you mean.
By the way that clearly does not converge.

• Calc 2 -

This is how it looks: hopefully this clarifies it a bit.

n^n/(4)^1+3*n

• Calc 2 -

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^(n-1)
which gets very large indeed

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