Proving Infinity of a Sequence: a(n)=[n+7]/[2+sin(n)]

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SUMMARY

The sequence defined by a(n) = (n + 7) / (2 + sin(n)) tends to infinity as n approaches infinity. L'Hôpital's Rule is not applicable in this case since the denominator does not approach zero or infinity. Instead, by establishing that 1 ≤ 2 + sin(n) ≤ 3 for all n, one can derive both lower and upper bounds for the sequence. As n becomes large, both bounds converge to infinity, confirming that a(n) indeed tends to infinity.

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Homework Statement



Prove that the following sequence (a(n)) has the property that a(n) tends to infinity as n tends to infinity.


Homework Equations


a(n)=[n+7]/[2+sin(n)]


The Attempt at a Solution



i tried l'hospital's rule, so i got 1/cos(n)...which wouldn't work.
so I am not sure how to do this question.
 
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L'Hopital's Rule doesn't apply here. The quotient has to be of the form 0/0 or infinity/infinity for L'H's Rule to be used.

Instead, use the fact that 1 <= 2 + sin(n) <= 3, for all n to get lower and upper limits on the value of (n + 7)/(2 + sin(n)), and then show that the lower and upper limit expressions both have the same limit as n gets large, thereby trapping the expression in the middle.
 


Since n+7 is unbounded, to apply l'Hopital to this you would need that the denominator is also unbounded (so you have an infinity/infinity form). It's not. The denominator is bounded and positive. That should tell you the limit right there.
 

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