Posted by Paul 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 
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. 
Calculus 
Paul,
what does Tn stand for?

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