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Show the series an=n^2-n is strictly monotonic if n>=1?
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You need to test if the given sequence is strictly increasing or decreasing, hence, you need to evaluate the difference of two consecutive members of the sequence, such that:
`a_(n+1) - a_n = (n+1)^2 - (n+1) - n^2 + n`
Raising to square yields:
`a_(n+1) - a_n = n^2 + 2n + 1 - n - 1 - n^2 + n`
Reducing dupliuacte members yields:
`a_(n+1) - a_n = 2n`
Since `n>=1` yields that `2n >= 2 > 0` , hence` a_(n+1) - a_n > 0` .
Hence, testing if the given sequence strictly increases or decreases yields that `a_(n+1) - a_n > 0 => a_(n+1) > a_n` , thus the sequence strictly increases.
Posted by sciencesolve on July 2, 2013 at 5:01 PM (Answer #1)
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