Question

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

Answer 1

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.

Answer 2

so you use the p series test?

Answer 3

well its like two different types put together - the denominator is a p-series

Answer 4

thanks!

Answer 5

what series test do you use for \[\sum_{n=1}^{\infty} (n-1)/(n4^n)\]

Answer 6

You can break this one up into two parts...\[(n/n4^{n})-(1/n4^{n})\]
In both series, the terms will approach zero.

Answer 7

Alright so you can break it up, but how do you tell if it converges or diverges using the various series tests?

Answer 8

The first part simplifies to a geometric series\[(1/4)^{n}\]

Answer 9

How did you get that to reduce to \[1/4^{n}\]

Answer 10

because the n's cancel

Answer 11

but the top is subtraction not multiplication? so you cant just cancel like that...

Answer 12

I broke up the fraction into two parts because you have (n-1) in the numerator.

Answer 13

\[(n-1)/(n4^{n}) = (n/n4^{n}) - (1/n4^{n})\]

Answer 14

i still dont understand what you all cancel out to get the (1/4)^n?

Answer 15

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.

Answer 16

so you ignore the (1/n4^n)?

Answer 17

You can't ignore it, but you can treat it as a separate series.

Answer 18

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?

Answer 19

I guess since you have a fraction you could try limit comparison.

Answer 20

alright thanks