1. ## convergence and differgence

find out whether this is divergent, c.convergent or abs convergent

$\displaystyle \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}$

and

$\displaystyle \sum sin^{2}\frac{1}{n}$

so I've basically ruled out the c.convergence..

but i havent the slightest idea how to start the first one?

for the second I thought that 1/n is divergent, which means the the whole series would be divergent..?

2. Originally Posted by akhayoon
find out whether this is divergent, c.convergent or abs convergent

$\displaystyle \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}$
Consider:

$\displaystyle \frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}=\frac{4^n[(1/2)^n-1]}{4^{n+1}[(3/4)^{n+1}+1]}=\frac{(1/2)^n-1}{4[(3/4)^{n+1}+1]}$

So the limit of the general term as $\displaystyle n \to \infty$ is $\displaystyle -1/4$, so the series diverges.

RonL

3. Originally Posted by akhayoon
might wanna fix that syntax error lol
he is just using the test for divergence. as $\displaystyle n \to \infty$ the general term goes to $\displaystyle - \frac 14 \ne 0$

4. [quote=CaptainBlack;133801]
Originally Posted by akhayoon
find out whether this is divergent, c.convergent or abs convergent

$\displaystyle \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}$

/quote]

Consider:

$\displaystyle \frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}=\frac{4^n[(1/2)^n-1]}{4^{n+1}[(3/4)^{n+1}+1]}=\frac{(1/2)^n-1}{4[(3/4)^{n+1}+1]}$

So the limit of the general term as $\displaystyle n \to \infnty$ is $\displaystyle -1/4$, so the series diverges.

RonL

so i don't understand how you reached -1/4...how did you get rid of all the rest..confused

5. Originally Posted by akhayoon
so i don't understand how you reached -1/4...how did you get rid of all the rest..confused
$\displaystyle (1/2)^n$ and $\displaystyle (3/4)^{n + 1}$ both go to zero as $\displaystyle n \to \infty$. they are just convergent geometric series. what is left if you make those two zero, is -1/4

6. Originally Posted by akhayoon
$\displaystyle \sum sin^{2}\frac{1}{n}$

$\displaystyle \sum \left(\sin\left[\frac{1}{n}\right]\right)^2$

converges as $\displaystyle \sin(x) \sim x$ as $\displaystyle x \to 0$, so you can trap $\displaystyle ((\sin(1/n))^2$ between positive multiples of $\displaystyle (1/x)^2$ and we know that $\displaystyle \sum_{n=1}^{\infty}1/n^2$ converges and so the given series converges by the comparison test (among other means).

RonL

RonL