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
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.
what does Tn stand for?
nth Term of sequence
I thought it would be clear from the context and the obvious use of the ratio test.
To determine whether the series in question is convergent or divergent, we can use the Limit Comparison Test.
First, let's rewrite the series expression as a fraction:
sum n=1 to infinity [(5n)^(3n)] / [(5^n + 3)^n]
Now, to apply the Limit Comparison Test, we need to find another series that we already know the convergence behavior of.
Let's consider the series sum n=1 to infinity (5^n)^n.
If we compare the terms of the two series, we can see that (5n)^(3n) / [(5^n + 3)^n] is less than or equal to (5^n)^n / [(5^n + 3)^n].
Now, let's take the limit of the ratio of these two series as n approaches infinity:
lim n->inf [(5n)^(3n) / [(5^n + 3)^n]] / [(5^n)^n / [(5^n + 3)^n]]
Simplified, this is:
lim n->inf [(5n)^(3n) / (5^n + 3)^n] * [(5^n + 3)^n / (5^n)^n]
Now, let's take the limit of each factor separately:
lim n->inf [(5n)^(3n) / (5^n + 3)^n] = 1 (by applying the limit laws and noticing that the numerator and denominator have the same degree of n)
lim n->inf [(5^n + 3)^n / (5^n)^n] = 1 (by applying the limit laws and simplifying)
Therefore, the limit of the ratio is 1.
Now, the Limit Comparison Test states that if the limit of the ratio is a positive finite number, the two series share the same convergence behavior.
Since our limit is 1, which is a positive finite number, we can conclude that the original series sum n=1 to infinity [(5n)^(3n)] / [(5^n + 3)^n] has the same convergence behavior as the series sum n=1 to infinity (5^n)^n.
The series sum n=1 to infinity (5^n)^n is a divergent series, as it grows exponentially. Therefore, by the Limit Comparison Test, the original series sum n=1 to infinity [(5n)^(3n)] / [(5^n + 3)^n] is also divergent.