2013-02-10, 07:26 PM
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)
![[Image: 2%7D=1]](http://latex.codecogs.com/gif.latex?\lim_%7Bn\to\infty%7D\frac%7B\frac%7B\sqrt_%7Bn%7D%7D%7Bn%5E2-3%7D%7D%7B\frac%7B1%7D%7Bn%5E%7B3/2%7D%7D%7D=\lim_%7Bn\to\infty%7D\frac%7B\sqrt_%7Bn%7D%7D%7Bn%5E2-3%7Dn%5E%7B3/2%7D=1)
Since
![[Image: 2%7D%7D]](http://latex.codecogs.com/gif.latex?\sum_%7Bn=2%7D%5E%7B\infty%7D\frac%7B1%7D%7Bn%5E%7B3/2%7D%7D)
converges, (p-series p = 3/2)
therefore
![[Image: gif.latex?\sum_%7Bn=2%7D%5E%7B\infty%7D\frac%7B\sqrt_%7Bn%7D%7D%7Bn%5E2-3%7D%7D]](http://latex.codecogs.com/gif.latex?\sum_%7Bn=2%7D%5E%7B\infty%7D\frac%7B\sqrt_%7Bn%7D%7D%7Bn%5E2-3%7D%7D)
converges
These images work, right?
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?