Question

prove divergence of the series (1+sqrt(j))/(j(ln(j))^2) from j=2 to infinity

Answer 1

Can you check the question?
If I interpret it as:
a(n)=(sqrt(n)+1)/(n(ln(n)^2))
then the limit of a(n+1)/a(n)=0 (ratio test)
so the series is absolutely convergent.

You can look at it in another way:
even sqrt(j)/j -> 0 as j-> inf.
so dividing in addition by ln(j)^2 can only make it disappear faster.

Note j starts from 2 because log(1)=0, and the term j=1 is undefined.

Answer 2

you're right, it says prove convergence or divergence, I'm sorry. But how did you get the limit of the ratio = 0?

Answer 3

The ratio test actually gives 1 as a limit, so it is inconclusive.
However, the term a(n) -> 0 as n->0 (necessary but not sufficient for convergence).
By expanding the numerator and denominator as a power series, we find that the difference of the highest power is only one. So it diverges by the power series rule.
Sorry for the misinformation earlier.

Answer 4

I don't think that I've learned that test,is there another way to show it?

Answer 5

Another way is to compare the sum of the series with the integral, but the equivalent function integrates to an incomplete gamma function, with which I am not familiar, that's why I did not suggest it.
A smarter way is to modify the function to be the lower bound of the series. If the modified function integrates to infinity, then the series diverges. I have not found one such function yet. Usually it should be one that is easier to integrate, and is a lower bound of the original integral.
Sorry, the only test that comes to mind for now is expanding the function to power series, apart from working out the integral test.

Answer 6

Try a new post and see if someone else has some fresh ideas!

Answer 7

alright, thank you! I have been working on this problem for hours and I can't seem to get it

Answer 8

Just some thoughts.
If we split the series into two according to the numerator, we have
S1=sum(1/(jln(j)^2)), and
S2=sum(sqrt(j)/(jln(j)^2)
I1=integral(dj/(jln(j)^2)), from 2 to inf, and
I2=integral(sqrt(j)dj/(jln(j)^2) from 2 to inf
To prove divergence, we need to prove that at least one of the two sums OR integrals is divergent.
We can use the substitution u=log(j), then du = dj/j, or
I1=int(du/u^2)
which easily integrates to ln(2) from 2 to inf.
I2 = int(sqrt(x)du/u^2)
=int(e^(u/2)/u^2) from 2 to inf
which is clearly divergent since the limit of e^(u/2)/u^2 is inf.
Sorry it took a little while, but this should do it.