1. ## Convergence/Divergence check

Not sure how to test this one:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}$

2. Originally Posted by cinder
Not sure how to test this one:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}$
$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}=-4~ e^{-1}$

so I guess its convergent.

Alternativly as it is an alternating series of tems whose absolute values are
decreasing and whose limit is 0, then it is convergent. Also it is absolutely
convergent as:

$\displaystyle 1/n! < n^2,~ \forall~ n>3$

and

$\displaystyle \sum_1^{\infty}1/n^2$

converges

RonL

3. I think it can be tested with the alternating series test, but I'm not sure how to go about it. The book says it converges if $\displaystyle 0 < a_{n+1} \leq a_n$ and $\displaystyle \lim_{n\rightarrow\infty} a_n=0$, but I don't understand that.

4. And is $\displaystyle \sum_{n=0}^{\infty} e^{-n}$ divergent?

5. Originally Posted by cinder
And is $\displaystyle \sum_{n=0}^{\infty} e^{-n}$ divergent?
Its a geometric series with common ratio less than 1, so yes it does converge.

RonL

6. Originally Posted by cinder
Not sure how to test this one:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n-1}4}{n!}$
Originally Posted by cinder
I think it can be tested with the alternating series test, but I'm not sure how to go about it. The book says it converges if $\displaystyle 0 < a_{n+1} \leq a_n$ and $\displaystyle \lim_{n\rightarrow\infty} a_n=0$, but I don't understand that.
Either my definition of $\displaystyle a_n$ from the series is wrong or the theorem should be:
$\displaystyle 0 < |a_{n + 1}| \leq |a_n|$

So:
$\displaystyle |a_n| = \frac{4}{n!}$

and

$\displaystyle |a_{n + 1}| = \frac{4}{(n + 1)!}$

We obviously have that
$\displaystyle 0 < |a_{n + 1}| \leq |a_n|$
so that part of the theorem is satisfied.

Now for
$\displaystyle \lim_{n \to \infty} \frac{(-1)^{n-1}4}{n!}$

The constant term, of course, is immaterial here. The (-1) term simply oscillates between -1 and 1, so if the other part of the limit is zero then we're okay. And, in fact $\displaystyle \lim_{n \to \infty}\frac{1}{n!} = 0$, so
$\displaystyle \lim_{n \to \infty} \frac{(-1)^{n-1}4}{n!} = 0$
as required.

-Dan