dirtypyjamas
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- May 16, 2013
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an=(-n^3+5)/(n(n^3+7n)^(1/2))
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Does this converge or diverge and how?
- Thread starter dirtypyjamas
- Start date
If it is the sequence you are considering: the trick is to make denominators or numerators larger or smaller to be convenient for you. Here I make the denominator larger, and hence the entire fraction smaller. Can you simplify and take a guess?
\(\displaystyle \dfrac{|-n^3+5|}{n\sqrt{n^3+7n}} \ge \dfrac{|-n^3+5|}{n\sqrt{n^3+15n^3}}\)
\(\displaystyle \dfrac{|-n^3+5|}{n\sqrt{n^3+7n}} \ge \dfrac{|-n^3+5|}{n\sqrt{n^3+15n^3}}\)
limit as n --> oo of an=(-n^3+5)/(n(n^3+7n)^(1/2))
dirtypyjamas,
also, as n approaches oo, the ratio of highest degrees of the numerator and denominator is what counts.
\(\displaystyle \dfrac{-n^3}{n(n^3)^{1/2}} \ = \)
\(\displaystyle \dfrac{-n^3}{n(n^{3/2})} \ = \)
\(\displaystyle \dfrac{-n^3}{n^{5/2}} \ = \)
\(\displaystyle \dfrac{-n^{6/2}}{n^{5/2}} \ = \)
\(\displaystyle {-n^{1/2}} \ \ \ (or \ \ -\sqrt{n}) \)
And then the \(\displaystyle \ \displaystyle\lim_{n \to \infty} ({-n^{1/2}}) \ \ diverges \ \ to \ \ -\infty.\)