determine absolute or conditional convergence:
sum(k=1 - infinity) [(-1)^(k+1)(10^k)]/(k!)

Question

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

anonymous · Answer

have u tried ratio test?

anonymous · Answer

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

anonymous · Answer

of course alternating series test will work :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

what is ur difficulty with that??

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

*increases

anonymous · Answer

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

anonymous · Answer

yes

anonymous · Answer

how to tell weather it converges absolutely or conditionally?

anonymous · Answer

emm...u should apply Absolute Convergence test

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

therefore it converges

anonymous · Answer

@mukushla

anonymous · Answer

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

anonymous · Answer

it does not*

anonymous · Answer

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.