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+--- Thread: Proving the convegence of the series sqrt(n)/(n^2 - 3) (/showthread.php?tid=62384)
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Imagine - 2013-02-10
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?
Proving the convegence of the series sqrt(n)/(n^2 - 3) - JoeTang - 2013-02-10
Why doesn't the ratio test work?
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Imagine - 2013-02-10
JoeTang Wrote:Why doesn't the ratio test work?
You end up with 1.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - JoeTang - 2013-02-10
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?
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Kalovale - 2013-02-11
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.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Stereo - 2013-02-12
JoeTang Wrote: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.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Declaimed - 2013-02-14
What level math is this? I don't think I've ever seen this before.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Stereo - 2013-02-14
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.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Imagine - 2013-02-14
Declaimed Wrote:What level math is this? I don't think I've ever seen this before.
Calculus C/first year college math for most people.
Proving the convegence of the series sqrt(n)/(n^2 - 3) - Locked - 2013-02-17
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.