anonymous
  • anonymous
determine absolute or conditional convergence: sum(k=1 -> infinity) [(-1)^(k+1)(10^k)]/(k!)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
have u tried ratio test?
anonymous
  • anonymous
well it's an alternating series. i think i'm supposed to use the alternating series test
anonymous
  • anonymous
of course alternating series test will work :)

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anonymous
  • anonymous
i'm not sure how i'm supposed to do it because of the k+1 on the -1
anonymous
  • anonymous
u just need to drop \[(-1)^{k+1}\]and then work on\[\frac{10^k}{k!}\]
anonymous
  • anonymous
this is a good sourse http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx
anonymous
  • anonymous
what is ur difficulty with that??
anonymous
  • anonymous
ok.. now i get that.. so how should i take the limit of 10^k/k! ?
anonymous
  • anonymous
oh sorry i was out ;) well k! is very greater than 10^k when k increses
anonymous
  • anonymous
*increases
anonymous
  • anonymous
so would that mean that the limit would go to zero?
anonymous
  • anonymous
yes
anonymous
  • anonymous
how to tell weather it converges absolutely or conditionally?
anonymous
  • anonymous
emm...u should apply Absolute Convergence test
anonymous
  • anonymous
sorry.. i kinda don't understand how to do that here... could you help me out??
anonymous
  • anonymous
ok u got the first part right? alternating series test?
anonymous
  • anonymous
yes, lim(k>infinity)[10^k/k!] = 0
anonymous
  • anonymous
therefore it converges
anonymous
  • anonymous
@mukushla
anonymous
  • 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.
anonymous
  • anonymous
it does not*
anonymous
  • 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.

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