It is divergent. The numerator is quadradic while denominator is cubic, their ratio is simpler behaving like 1/k which is divergent.
Determine whether the series is convergent or divergent. If it is convergent, find it's sum.
It seems to me that it should converge, because the power of the denominator is greater than the power of the numerator, but I'm having a tough time figuring out how to find the answer.
Thank you, would this be an acceptable answer?
If we evaluate a given term in our sum as it moves towards infinity, we get
(Note that I'm not sure if this step is valid, it seems unlikely that I can evaluate some of the limits while not evaluating others, but if I can, then I think my answer will work)
So we can see that as k moves towards infinity, our terms converge to the same values as the terms in the sum
Which is a divergent series known as the Harmonic series.
Thus our equation is divergent.
Sorry, but all you've really done here is show that . This is a necessary but not sufficient condition for convergence.
You need to use the comparison test:
.
So .
But diverges. Therefore ......
The tricks used in increasing the denominator and decreasing the numerator to get a simple comparison are handy ones to remember.
You'd use them in the 'opposite way' if you suspected convergence and therefore were trying to show that a_k < b_k where sum b_k converges.
But I think the best divergence/convergence test is the integral test... if you rewrite which gives you + evaluated from 0-∞ and as you can see without even evaluating it evaluated at ∞ gives ∞ therefore the integral converges...ergo the summation diverges.