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anonymous
 5 years ago
I wanted to know why I can't apply the alternating series test to this series sigma n=1 to infinity (1)^n1 sin n
anonymous
 5 years ago
I wanted to know why I can't apply the alternating series test to this series sigma n=1 to infinity (1)^n1 sin n

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you make the expression clearer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it,\[()^{n1}\sin n\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty}(1)^{n1} \sin( n )\] maybe that's what he meant?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, that's what I've read :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah thats what it is ty

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The alternating series test says you need the following for it to apply:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your 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
 5 years ago
Best ResponseYou've already chosen the best response.0sin(n) does *not* monotonically decrease to zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't apply the test here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u mean when we plug in values in the series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. A you increase n, you should get a decrease in b_n...and this shouldn't stop.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But sin(n) will decrease and then start to increase again, and then start to decrease and then...as n increases.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but when i plug in values it seems like its decreasing 0,.909,.141,..756

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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^%28n%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Here, 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
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0in the example above are we able to use it ∑n=1∞cosnπ/n2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, you can use it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The bit that isn't alternating needs to do two things: 1. go to zero as n goes to infinity 2. be monotonic decreasing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There'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.
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