anonymous
  • anonymous
What series test do you use when the sum starts at n =1 (2 + (-1)^n)/(n(n^(1/2)))?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
In the numerator you'll end up with 2+1, 2-1, 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
  • anonymous
so you use the p series test?
anonymous
  • anonymous
well its like two different types put together - the denominator is a p-series

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anonymous
  • anonymous
thanks!
anonymous
  • anonymous
what series test do you use for \[\sum_{n=1}^{\infty} (n-1)/(n4^n)\]
anonymous
  • anonymous
You can break this one up into two parts...\[(n/n4^{n})-(1/n4^{n})\] In both series, the terms will approach zero.
anonymous
  • anonymous
Alright so you can break it up, but how do you tell if it converges or diverges using the various series tests?
anonymous
  • anonymous
The first part simplifies to a geometric series\[(1/4)^{n}\]
anonymous
  • anonymous
How did you get that to reduce to \[1/4^{n}\]
anonymous
  • anonymous
because the n's cancel
anonymous
  • anonymous
but the top is subtraction not multiplication? so you cant just cancel like that...
anonymous
  • anonymous
I broke up the fraction into two parts because you have (n-1) in the numerator.
anonymous
  • anonymous
\[(n-1)/(n4^{n}) = (n/n4^{n}) - (1/n4^{n})\]
anonymous
  • anonymous
i still dont understand what you all cancel out to get the (1/4)^n?
anonymous
  • anonymous
The 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
  • anonymous
so you ignore the (1/n4^n)?
anonymous
  • anonymous
You can't ignore it, but you can treat it as a separate series.
anonymous
  • anonymous
ok 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
  • anonymous
I guess since you have a fraction you could try limit comparison.
anonymous
  • anonymous
alright thanks

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