Posted by **Paul** on Friday, September 21, 2012 at 12:47pm.

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 -
**Steve**, Friday, September 21, 2012 at 3:53pm
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 -
**Paul**, Friday, September 21, 2012 at 4:41pm
what does Tn stand for?

- Calculus -
**Steve**, Saturday, September 22, 2012 at 12:13am
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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