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