1. Need Help with Series!

hey guys, really need help with calculus... its my last class i need to pass in order to gaduate and get
the dammed bachelors degree. any who...

the problem is that i dont understand anything i started doing the homework i need to finish today .. and got stuck on the first exercise here it is..

series of (3/n^2 - 16) n = 0 to infinity
$\displaystyle \sum^{\infty}_{n=0} ({\frac{{\frac{3}{(n+1)^2 -16}}}{{\frac{3}{n^2 - 16}}}})$

ok i know this is a Series P and P = 2 therefore is greater than 1 making this series converge

thing is i need to apply the ratio formula.. wich is :

limit(3/(((n+1)^2-16)*(3/(n^2-16))), n = infinity)

maple says the limit is 1, but when i do it manually i get -(12/15)

what am i doing wrong??

here are my steps

3
---------
(n+1)^2 -16
-----------
3
---------
n^2 - 16

ok so ...the 3's cancel each other out and you get

lim (n+1)^2 -16
----------
n^2 - 16

now you make a substitution, n = 1

lim (1+1) ^2 - 16
--------------
1^2 - 16

that turns into

lim 4 - 16
------
1 - 16

that equals

- 12/15

now... -12/15 is not 1....

what am i doing wrong?? thx in advance!!

2. I turned it into $\displaystyle a_n=\frac{n^2-16}{(n+1)^2-16}$

From here it's apparent that $\displaystyle a_n \rightarrow 1$ (and therefore the series does not converge since $\displaystyle a_n \rightarrow 0$ for that to happen) but you need the ratio test.

Hence:

$\displaystyle |\frac{a_{n+1}}{a_n}|=\frac{((n+1)^2-16)((n+1)^2-16)}{((n+2)^2-16)(n^2-16)} \rightarrow 1$ (you can expand all this out if you want, it isn't apparent it converges to 1 in this form).

I don't know why you substituted 1 into the equation.

Is this what you needed?

3. Showcase_22, thank you so much for your fast reply, basically i need to know whether it converges or not using the ratio, direct comparison and the integral test.

and i needed to know how the limit of
$\displaystyle a_n=\frac{n^2-16}{(n+1)^2-16}$

got to 1, since in class the profeesor told us that if the lim was 1 it was inconclusive but if the limit as less than 1 it converged and if it was greater than 1 then it diverged.

4. With the ratio test, it's $\displaystyle |\frac{a_{n+1}}{a_n}| \rightarrow L$ as $\displaystyle n \rightarrow \infty$ not as $\displaystyle n \rightarrow 1$.

Since the ratio test tends to 1, we can't use it here (as you rightly said).

The series does not converge. A necessary condition for it to converge is that $\displaystyle a_n \rightarrow 0$ but it doesn't. Isn't this enough?

If not, try and make $\displaystyle \sigma a_n$ larger than something that diverges.