Proving the convegence of the series sqrt(n)/(n^2 - 3)

[Imagine]

I have a series sqrt(n)/(n^2 - 3) that goes from 2 to infinity, and I want to see if it converges or diverges. I tried the ratio test, the test for divergence, and the comparison test, but none of them work (I didn't use the integral test because the teacher will not have a problem where the integral is complex for most people to do). Is there any way for me to prove that the series converges or diverges?

[JoeTang]

Why doesn't the ratio test work?

[Imagine]

Why doesn't the ratio test work?

You end up with 1.

[JoeTang]

My bad. Brain fart.

You should be using the limit comparison test:

which states lim(n->infinity) a_n/b_n = c, 0 < c < infinity
where a_n and b_n >= 0;

the series a_n converges if b_n converges

For very large n, you can approximate
sqrt(n)/(n^2-3) ~= sqrt(n)/(n^2) = 1/n^3/2 = b_n
(as n approaches infinity, -3's effect approaches 0)



Since

converges, (p-series p = 3/2)

therefore

converges

These images work, right?

[Kalovale]

You only have so many tools for these problems. There was only one problem that I had to get creative with, and it involved a limit comparison test. I still have no idea how I thought of that b_n, but it turned out to work.

[Stereo]

For very large n, you can approximate
sqrt(n)/(n^2-3) ~= sqrt(n)/(n^2) = 1/n^3/2 = b_n
(as n approaches infinity, -3's effect approaches 0)

If you don't want to be like "for very large n", you can also observe that
sqrt(n)/(n^2-3) < sqrt(n)/(n^2/2)
for n>2 this holds, so you can squeeze the terms under 1/2*n^(-3/2) which is half a convergent p-series.
n=1 and n=2 are not a problem cause there are only a finite number of them, and none are infinite.


It'd be simpler on something like sqrt(n)/(n^2+2) because you can directly observe
sqrt(n)/(n^2+2) < sqrt(n)/(n^2)
for any n>0.

[Declaimed]

What level math is this? I don't think I've ever seen this before.

[Stereo]

Limits are usually introduced before calculus, as understanding them is part of the calculus fundamentals. So it could be 12th grade or first year university, where I live.

How detailed it gets probably depends on the audience of the course, should at least see them by calc though.

[Imagine]

What level math is this? I don't think I've ever seen this before.

Calculus C/first year college math for most people.

[Locked]

Not all first years take Calculus 2, most don't even get to that level. Some will stop at Calculus 1, others begin with college algebra. I think you give people too much credit in their mathematical "prowess". It's hard to find people who enjoy math enough to take it to a higher level besides those that are required such as engineering.