1. ## Sequences Questions (2)

Two questions for y'all; both involve determining convergence or divergence, if convergence finding limit:

1) a(sub n) = SQRT((n+1)/(9n+1)): I see that you can make that (N+1)^1/2 divided by (9n + 1)^1/2, but can't progress.

2) a(sub n) = ((-1)^n*n^3)/(n^3+2n^2+1): Since there is the -1^n term, is this simply divergent? Or am I looking at it the wrong way?

Cheers!

2. Originally Posted by Sprintz
Two questions for y'all; both involve determining convergence or divergence, if convergence finding limit:

1) a(sub n) = SQRT((n+1)/(9n+1)): I see that you can make that (N+1)^1/2 divided by (9n + 1)^1/2, but can't progress.

2) a(sub n) = ((-1)^n*n^3)/(n^3+2n^2+1): Since there is the -1^n term, is this simply divergent? Or am I looking at it the wrong way?

Cheers!

1) $a_n=\sqrt{\frac{n+1}{9n+1}}$

First lets factor out the highest power of n from both the numerator and denominator to get

$a_n=\sqrt{\frac{n(1+\frac{1}{n})}{n(9+\frac{1}{n}) }}$

and reduce to get

$a_n=\sqrt{\frac{(1+\frac{1}{n})}{(9+\frac{1}{n})}}$

Now as $n \to \infty$ $\frac{1}{n} \to 0$ so we get

$a_n=\sqrt{\frac{(1+0)}{(9+0)}}=\frac{1}{3}$

Your are correct that the 2nd sequence diverges, but try an analysis similar to the one I just did. Good luck.