Asked by Paul

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

Answers

Answered by Steve
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.
Answered by Paul
what does Tn stand for?
Answered by Steve
nth Term of sequence
I thought it would be clear from the context and the obvious use of the ratio test.
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