## anonymous 4 years ago determine absolute or conditional convergence: sum(k=1 -> infinity) [(-1)^(k+1)(10^k)]/(k!)

1. anonymous

have u tried ratio test?

2. anonymous

well it's an alternating series. i think i'm supposed to use the alternating series test

3. anonymous

of course alternating series test will work :)

4. anonymous

i'm not sure how i'm supposed to do it because of the k+1 on the -1

5. anonymous

u just need to drop $(-1)^{k+1}$and then work on$\frac{10^k}{k!}$

6. anonymous

this is a good sourse http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

7. anonymous

what is ur difficulty with that??

8. anonymous

ok.. now i get that.. so how should i take the limit of 10^k/k! ?

9. anonymous

oh sorry i was out ;) well k! is very greater than 10^k when k increses

10. anonymous

*increases

11. anonymous

so would that mean that the limit would go to zero?

12. anonymous

yes

13. anonymous

how to tell weather it converges absolutely or conditionally?

14. anonymous

emm...u should apply Absolute Convergence test

15. anonymous

sorry.. i kinda don't understand how to do that here... could you help me out??

16. anonymous

ok u got the first part right? alternating series test?

17. anonymous

yes, lim(k>infinity)[10^k/k!] = 0

18. anonymous

therefore it converges

19. anonymous

@mukushla

20. anonymous

$\lim_{k \rightarrow \infty}a_k=0$ Is a REQUIREMENT for a series to converge but it doesn't not IMPLY that a series converges.

21. anonymous

it does not*

22. anonymous

Even if it is an alternating series do the ratio test, the ratio test works well for factorials. Also, if that converges then you know it absolutely converges (because you take the absolute value so the (-1)^k+1 goes away) and if it absolutely converges then you know it conditionally converges.