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 2 years ago
Here's a painful mix of real analysis and caluclus ;D
Does the following series converge or diverge?
http://data.artofproblemsolving.com/images/latex/1/3/8/13826d69351e71414333a1b1ad64f591c4a9eb2a.gif
As taken from the Art of Problem Solving.
 2 years ago
Here's a painful mix of real analysis and caluclus ;D Does the following series converge or diverge? http://data.artofproblemsolving.com/images/latex/1/3/8/13826d69351e71414333a1b1ad64f591c4a9eb2a.gif As taken from the Art of Problem Solving.

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badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1Lol, I don't understand the solution myself. But I thought people here would appreciate it. Right now, I'm trying to figure it out.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1\[\begin{align} \sum_{n=1}^{\infty}\frac{(\sin(n)+2)^n}{3^nn}&=\sum_{n=1}^{\infty}\frac{1}{n}\times\frac{(\sin(n)+2)^n}{3^n}\\ &=\sum_{n=1}^{\infty}\frac{1}{n}\times(\frac{\sin(n)+2}{3})^n\\ \end{align}\]The maximum value \(\sin(n)\) is 1, therefore the maximum value of the fraction \(\frac{\sin(n)+2}{3}=\frac{1+2}{3}=1\). Therefore the original sum will include a subsum involving a sequence of \(\frac{1}{n}\) for every \(n\) that gives \(\sin(n)=1\), and, I believe, this subsequence is divergent, therefore the original sequence is divergent.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I'm confused about one thing asnaseer didn't you show that\[\frac{1}{n}\times(\frac{\sin(n)+2}{3})^n<\frac1n\]so how does the fact that the sum of 1/n diverges help us if it is larger than the function in question?

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1It is less, but I believe, for an infinite sum, taking out some intermediate terms from the sum should not affect its convergent/divergent behaviour. I am not an expert in this field and may, therefore, have missed some important point here. I am just working off "gut instincts" :)

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1If you guys are interested in seeing the solution, it's here. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=70&t=22093 I'm having difficulty following Hata's theorem.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1It converges, by the way.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1The discussion on that link does not seem to conclusively prove convergence. Also, I tried wolfram and it says that the sum is divergent: http://www.wolframalpha.com/input/?i=sum+%28sin%28n%29%2B2%29%5En%2F%28n*3%5En%29+for+n%3D1+to+infinity I guess this needs someone with more experience in real analysis to point us in the correct direction for proving this one way or another. Maybe Zarkon, JamesJ or FoolForMath can help?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1You're right, of course. Also, I have no idea what Hata's theorem is. Google is useless on this issue. >.>

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1Yes  that is why I am hoping one of the big guns on this site sees this question and points us in the right direction. :)

dave444
 2 years ago
Best ResponseYou've already chosen the best response.2The problem with the subseries method outlined above to prove divergence, is we cannot be sure there are an infinite number of n such that sin(n)=1. Are there in fact any? For sin(x)=1, x = pi/2 + 2k pi, for integer k. I don't think there is any n so that pi/2+2k pi = n for n and k both integers. (Otherwise, we could express pi as the ratio of 2 integers, which contradicts the fact that pi is irrational.)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1good point dave444, but I was interpreting the n in sin(n) as degrees  maybe that is the wrong interpretation?

dave444
 2 years ago
Best ResponseYou've already chosen the best response.2In problems of real analysis or calculus, one should always assume radians unless told otherwise.

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.1I see  thanks for the insight! :)

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.1Here's the solution as given by some really smart guys.
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