anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Can you make the expression clearer?
anonymous
  • anonymous
Is it,\[(-)^{n-1}\sin n\]?
anonymous
  • anonymous
\[\sum_{n=1}^{\infty}(-1)^{n-1} \sin( n )\] maybe that's what he meant?

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anonymous
  • anonymous
lol, that's what I've read :)
anonymous
  • anonymous
yah thats what it is ty
anonymous
  • anonymous
The alternating series test says you need the following for it to apply:
anonymous
  • 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
  • anonymous
sin(n) does *not* monotonically decrease to zero.
anonymous
  • anonymous
It oscillates.
anonymous
  • anonymous
You can't apply the test here.
anonymous
  • anonymous
u mean when we plug in values in the series
anonymous
  • anonymous
Yes. A you increase n, you should get a decrease in b_n...and this shouldn't stop.
anonymous
  • anonymous
But sin(n) will decrease and then start to increase again, and then start to decrease and then...as n increases.
anonymous
  • anonymous
but when i plug in values it seems like its decreasing 0,.909,.141,..756
anonymous
  • 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
  • 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
  • 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
  • anonymous
right it depends
anonymous
  • 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
  • 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
  • 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
  • anonymous
in the example above are we able to use it ∑n=1∞cosnπ/n2
anonymous
  • 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
  • anonymous
Oh yeah, you can use it.
anonymous
  • anonymous
okay ty
anonymous
  • 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
  • anonymous
*monotone
anonymous
  • anonymous
k ty
anonymous
  • 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
  • anonymous
cool ty for info
anonymous
  • anonymous
ok. good luck.

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