## privetek 3 years ago determine absolute or conditional convergence: sum(k=1 -> infinity) [(-1)^(k+1)(10^k)]/(k!)

1. mukushla

have u tried ratio test?

2. privetek

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

3. mukushla

of course alternating series test will work :)

4. privetek

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

5. mukushla

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

6. mukushla

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

7. mukushla

what is ur difficulty with that??

8. privetek

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

9. mukushla

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

10. mukushla

*increases

11. privetek

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

12. mukushla

yes

13. privetek

how to tell weather it converges absolutely or conditionally?

14. mukushla

emm...u should apply Absolute Convergence test

15. privetek

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

16. mukushla

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

17. privetek

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

18. privetek

therefore it converges

19. privetek

@mukushla

20. malevolence19

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

21. malevolence19

it does not*

22. malevolence19

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.