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anonymous
 5 years ago
Consider the following sequence: An = 2(1)^n + (n/(n+1)). Is it possible to prove that An <= 3 for every n? Thanks
anonymous
 5 years ago
Consider the following sequence: An = 2(1)^n + (n/(n+1)). Is it possible to prove that An <= 3 for every n? Thanks

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you only have to show that lim n/(n+1) =1 as n tends to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the justification for that assumption?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivate the function An with respect to n. Put the derivative equals to zero to find the value of n on the boundry. Then put the value of n back in An and evaluate n. This will always be less then or equal to 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is because the derivative equals to zero gives us the turning point of the function... ie either the value of n coresponding to greatest or least value of An. Take the double derivative and put the value of n in it, it will turn out to be ve showing this is the maximum point so all other values of An are less then this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well...the derivative is 1/((n+1)^2).....therefore, is always positive.... > 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what abt the derivative of 2*(1)^n?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont think so... its something involving log.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i jst checked the above reply of limit aproach to infinity... that wud work too as n/1+n will approach 1 at infinity... and 2*(1)^n can either be 2 or +2 so that wud gve correct answr

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understood the limit reply....but it is safe to assume that the limit of n/(n+1) as n tends to infinity plus 2 is the maximum term of An?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you don't need the derivative, and you certainly don't need the derivative of \[2(1)^n\] the biggest \[2(1)^n\] can be is 2 and \[\frac{n}{n+1}<1\] by inspection. the numerator is less that the denominator!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok....but in order to prove that An <= 3 for every n, shouldn't we prove that An = 3 (which is impossible)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand less than 3, but less or EQUAL than 3 is a little bit more complicated to understand...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it is less than three it is certainly less than or equal to three. and less than 4 too!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0perhaps i was being silly. \[0<3\] is a true statement, but so is \[0\leq3\] If you know something is less than three it is certainly less than or equal to 3. The sequence you had will never be 3. There will be values that get closer and closer to 3 (and to 1, so the sequence has no actual limit), but it will certainly never be 3.
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