Consider the following sequence: An = 2(-1)^n + (n/(n+1)). Is it possible to prove that An <= 3 for every n? Thanks
you only have to show that lim n/(n+1) =1 as n tends to infinity.
what is the justification for that assumption?
derivate 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
this 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
well...the derivative is 1/((n+1)^2).....therefore, is always positive.... > 0
what abt the derivative of 2*(-1)^n?
zero? :)
i dont think so... its something involving log.
i 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
i 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?
yes
you 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!
ok....but in order to prove that An <= 3 for every n, shouldn't we prove that An = 3 (which is impossible)?
i understand less than 3, but less or EQUAL than 3 is a little bit more complicated to understand...
if it is less than three it is certainly less than or equal to three. and less than 4 too!
perhaps 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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