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anonymous
 5 years ago
What series test do you use when the sum starts at n =1 (2 + (1)^n)/(n(n^(1/2)))?
anonymous
 5 years ago
What series test do you use when the sum starts at n =1 (2 + (1)^n)/(n(n^(1/2)))?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the numerator you'll end up with 2+1, 21, 2+1 etc. but in the denominator you have n^(3/2) which will go to infinity as n gets large. So the terms will go to zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you use the p series test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well its like two different types put together  the denominator is a pseries

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what series test do you use for \[\sum_{n=1}^{\infty} (n1)/(n4^n)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can break this one up into two parts...\[(n/n4^{n})(1/n4^{n})\] In both series, the terms will approach zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright so you can break it up, but how do you tell if it converges or diverges using the various series tests?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first part simplifies to a geometric series\[(1/4)^{n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get that to reduce to \[1/4^{n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the n's cancel

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the top is subtraction not multiplication? so you cant just cancel like that...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I broke up the fraction into two parts because you have (n1) in the numerator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(n1)/(n4^{n}) = (n/n4^{n})  (1/n4^{n})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i still dont understand what you all cancel out to get the (1/4)^n?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first part (n/n4^n)  the n's will cancel and you're left with (1/4^n) = (1/4)^n  which is a geometric series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you ignore the (1/n4^n)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't ignore it, but you can treat it as a separate series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so if im trying to find if the series is convergent or divergent... i represent with evidence using the geometric series as a whole? Because up til this section we are limited to the geometric, telescoping, limit comparison, direct comparison, p series and integral tests?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess since you have a fraction you could try limit comparison.
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