anonymous
  • anonymous
Determine whether the sequence converges or diverges. If it converges, nd the limit n!/2^n
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
You can use the ratio test. \[\lim_{n->\infty}\left|\frac{a_{n+1}}{a_n} \right|=\lim_{n->\infty}\frac{(n+1)!/2^{n+1}}{n!/2^{n}}=\lim_{n->\infty}\frac{(n+1)n!}{2 \times 2^n}\frac{2^n}{n!}\]\[=\lim_{n->\infty}\frac{n}{2} \rightarrow \infty\] The sequence does not converge since the ratio in the limit is not less than 1.
anonymous
  • anonymous
nice!
anonymous
  • anonymous
Thanks ;)

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anonymous
  • anonymous
\[\frac{n+1}{2}\]
anonymous
  • anonymous
A question.. So does it mean that you cant test some values as N... and see whats happening?
anonymous
  • anonymous
Can you rephrase that? I'm not sure what you mean.
anonymous
  • anonymous
n!/2^n... Put in some values, as a check.. ..n= 1, 2, 3
anonymous
  • anonymous
And see if it diverges or converges
anonymous
  • anonymous
I hope you didnt misunderstand me, it was just a question, if i would in some values, in the funktion , n!/2^n .. could i see right away if it diverges or converges..
anonymous
  • anonymous
Ah I see. No, because that's not actually testing for ALL n. The tests are derived from the definition of convergence.
anonymous
  • anonymous
You could get an idea, but it wouldn't prove it. That's all.
anonymous
  • anonymous
Okey, so u have to, show it for all n. like u did
anonymous
  • anonymous
Exactly.
anonymous
  • anonymous
Well, thx.
anonymous
  • anonymous
No worries.
anonymous
  • anonymous
Hehe, I thought this was XiaoHong's question.
anonymous
  • anonymous
So did i ;)
anonymous
  • anonymous
thanks
anonymous
  • anonymous
Welcome.

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