Post a New Question


posted by on .

I was able to solve (e), this is the only one I don't get.

Determine whether the series is convergent or divergent and say what test you used to solve it.

(d) sum n=1 to infinity
(5n)^(3n) / (5^n + 3)^n

  • Calculus - ,

    Tn = (5n)^(3n) / (5^n + 3)^n
    Tn+1 = (5(n+1))^(3(n+1)) / (5^(n+1) + 3)^(n)

    Tn+1/Tn = (5(n+1))^(3(n+1)) / (5^(n+1) + 3)^(n+1) * (5^n + 3)^n / (5n)^(3n)

    As n->oo, 5^n+3 is just 5^n, so we can simplify things a bit to

    (5n)^(3n) * (5n)^3 * 5^(n^2)
    ----------------------------- =
    (5n)^(3n) * 5^n * 5(n^2)

    (5n)^3 / 5^n

    Since exponentials grow faster than powers, the ratio is less than 1, so the series converges absolutely.

  • Calculus - ,

    what does Tn stand for?

  • Calculus - ,

    nth Term of sequence
    I thought it would be clear from the context and the obvious use of the ratio test.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question