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Calc 2

posted by on .

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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