anonymous
  • anonymous
Alternating Series. Determine if Sigma (n=1 to infinity) of (-1)^n (2n/15n+1) converges. I know that we take the limit of (2n/15n+1) as n approaches infinity. But; I'm not sure what to do since we get infinity/infinity when we take the limit...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
convergence looks at the ration of next/now to see how fast it goes to zero
amistre64
  • amistre64
but it helps if a read it right
amistre64
  • amistre64
we can determine this from the horizontal asymptote i believe

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amistre64
  • amistre64
2n/(15n+1) , for larger and larger value of n looks more like: 2n/15n would you agree?
anonymous
  • anonymous
Absolutely! !!
anonymous
  • anonymous
So now we have an (An) and (bn) to try the limit comparison test?
amistre64
  • amistre64
then for large value of n. this limits out to 2/15
amistre64
  • amistre64
we dont even need to do a limit comparison test since this thing doesnt even go to zero
amistre64
  • amistre64
if the limit of the function doesnt go to zero, it diverges
anonymous
  • anonymous
Oooo OK!
anonymous
  • anonymous
Could we also use L'Hopital's rule since we get an indeterminate form of infinity/infinity? Which would then give us that 2/15?
amistre64
  • amistre64
no need, but yes, you could if you got bored enough to carry it on :)
anonymous
  • anonymous
YES!
anonymous
  • anonymous
I think it might be a little different since it is an alternating sequence. Consider this.\[\sum_{n=1}^{\infty}(-1)^n (2n/(15n+1))=\sum_{k=1}^{\infty}(-\frac{4k-2}{30k-14}+\frac{4k}{30k+1})\]
anonymous
  • anonymous
The first term is the 2k-1 term, corresponding with the odd values of n (which are all negative), while the second term is the 2k term that corresponds to the even values of n.
anonymous
  • anonymous
Continuing....\[=\sum_{k=1}^{\infty})\frac{(2-4k)(30k+1)+4k(30k-14)}{(30k-14)(30k+1)}\]
anonymous
  • anonymous
\[=\sum_{k=1}^{\infty}\frac{-120k^2+56k+2+120k-56k}{900k^2-390k-14}=\sum_{k=1}^{\infty}\frac{2}{900k^2-390k-14}\]
anonymous
  • anonymous
Pretty sure that converges.
anonymous
  • anonymous
Certainly, if the series is not alternating, it diverges, but since it alternates it will converge.
anonymous
  • anonymous
@squirefreak Take a look at my work on your problem. I came to a different conclusion.

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