Question

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

Answer 1

Can you make the expression clearer?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yah thats what it is ty

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

It oscillates.

Answer 10

You can't apply the test here.

Answer 11

u mean when we plug in values in the series

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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,\[\sum_{n=1}^{\infty}\frac{\cos n \pi}{n^2}\]

Answer 18

right it depends

Answer 19

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.

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

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."

Answer 24

Oh yeah, you can use it.

Answer 25

okay ty

Answer 26

The bit that isn't alternating needs to do two things:

1. go to zero as n goes to infinity
2. be monotonic decreasing

Answer 27

*monotone

Answer 28

k ty

Answer 29

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.

Answer 30

cool ty for info

Answer 31

ok. good luck.