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

1. 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.

2. anonymous

so you use the p series test?

3. anonymous

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

4. anonymous

thanks!

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

How did you get that to reduce to $1/4^{n}$

10. anonymous

because the n's cancel

11. anonymous

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

12. anonymous

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

13. anonymous

$(n-1)/(n4^{n}) = (n/n4^{n}) - (1/n4^{n})$

14. anonymous

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

15. 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.

16. anonymous

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

17. anonymous

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

18. 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?

19. anonymous

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

20. anonymous

alright thanks