I wanted to know why I can't apply the alternating series test to this series sigma n=1 to infinity (-1)^n-1 sin n

- anonymous

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- anonymous

Can you make the expression clearer?

- anonymous

Is it,\[(-)^{n-1}\sin n\]?

- anonymous

\[\sum_{n=1}^{\infty}(-1)^{n-1} \sin( n )\] maybe that's what he meant?

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## More answers

- anonymous

lol, that's what I've read :)

- anonymous

yah thats what it is ty

- anonymous

The alternating series test says you need the following for it to apply:

- anonymous

Your terms need to be expressible in the form\[a_n=(-)^nb_n\]and the b_n have to be monotonic decreasing to zero as n approaches infinity.

- anonymous

sin(n) does *not* monotonically decrease to zero.

- anonymous

It oscillates.

- anonymous

You can't apply the test here.

- anonymous

u mean when we plug in values in the series

- anonymous

Yes. A you increase n, you should get a decrease in b_n...and this shouldn't stop.

- anonymous

But sin(n) will decrease and then start to increase again, and then start to decrease and then...as n increases.

- anonymous

but when i plug in values it seems like its decreasing 0,.909,.141,..756

- anonymous

This is what sine does for increasing n:
http://www.wolframalpha.com/input/?i=sin%28n%29
Compare this with e^(-n) which *is* monotonic decreasing:
http://www.wolframalpha.com/input/?i=e^%28-n%29

- anonymous

i understand it now but this is the the behavior of sin, so we can never use the alternative test bc its not always decreasing

- anonymous

Well, if it's like this, you can't use alternating series test. But don't be tricked into then thinking you can NEVER apply it whenever you see sine or cosine, because sometimes you can have series like,\[\sum_{n=1}^{\infty}\frac{\cos n \pi}{n^2}\]

- anonymous

right it depends

- anonymous

Here, you *can* use the test because, if you look at what happens as you increase n from 1 to 2 to 3 to...you see for cos(n pi):\[\cos(\pi)=-1\]\[\cos(2\pi)=1\]\[\cos(3 \pi)= -1\]in which case, the series can be written\[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\]This is alternating, and the term 1/n^2 is monotonic decreasing...so the test will apply.

- anonymous

i understand what u r saying , i have to see what i am given to determine whether ia m able to use it or not

- anonymous

Yes, always, and it's the same for any other test. Theorems in mathematics are derived from statements assumed at the beginning. These are your conditions. If the conditions don't exist, the theorem doesn't apply.

- anonymous

in the example above are we able to use it ∑n=1∞cosnπ/n2

- anonymous

When you embark on solving these problems, read through the assumptions for the test and check them off like a recipe..."Do I have monotone decreasing? Yes. Do I have blah blah? No, so can't use this theorem...have to look for something else."

- anonymous

Oh yeah, you can use it.

- anonymous

okay ty

- anonymous

The bit that isn't alternating needs to do two things:
1. go to zero as n goes to infinity
2. be monotonic decreasing

- anonymous

*monotone

- anonymous

k ty

- anonymous

There's a lot of information lying around. These notes are pretty good
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx
and there's stuff from MIT and the like either in open course ware or YouTube.

- anonymous

cool ty for info

- anonymous

ok. good luck.

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