study in monotonicaly`(n+a)/(n+5)` study also an+3/n+2

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embizze eNotes educator| Certified Educator

Assuming `n in NN,n>0` ; a sequence is monotonic if `a_1<=a_2<=a_3<=...` or decreasing.

(1) `n+a/n+5`

If a=1 the sequence is `7,7.5,,9.25,...` which is monotonically increasing.

If a=2 the sequence is `8,8,,9.5,...` which is monotonic. The terms of the sequence never decrease. It is not strictly monotonic.

If a=3 the sequence is `9,8.5,9,9.75,...` which is not monotonic.





The smallest value for n is 1, so `a<=2` .

The sequence is monotonically increasing for every `a<=2`

If `n+1+a/(n+1)+5<=n+a/n+5` we get `a>=n(n+1)` which is not true for any constant a, so the sequence can never be monotonically decreasing.

(2) `an+3/n+2:`






So this sequence is monotonically increasing for all `a>=3/2` .

If `a(n+1)+3/(n+1)+2<=an+3/n+2` we get `a<=3/(n(n+1))` which implies that a<0. Thus for any a<0 the sequence is monotonically decreasing.

sciencesolve eNotes educator| Certified Educator

You need to study if the given sequence increases or decreases, hence, you need to evaluate the difference of two consecutive terms of sequence and then you need to discuss the result, such that:

`((n+1)+a)/((n+1)+5) - (n+a)/(n+5)`

`(n+a+1)/(n+6) - (n+a)/(n+5)`

Bringing the terms to a comon denominator yields:

`((n+5)(n+a+1) - (n+a)(n+6))/((n+5)(n+6))`

Opening the brackets yields:

`(n^2 + na + n + 5n + 5a + 5 - n^2 - 6n - an - 6a)/((n+5)(n+6))`

Reducing like terms yields:

`(5 - a)/((n+5)(n+6))`

Since n is a natural number, hence, the denominator is positive, thus, you need to establish the sign of numerator such that:

If `a < 5 => 5-a>0 => ((n+1)+a)/((n+1)+5) - (n+a)/(n+5) > 0 =>`   the sequence increases

If `a> 5 => 5-a<0 => ((n+1)+a)/((n+1)+5) - (n+a)/(n+5)< 0 =>`  the sequence decreases

If `a = 5 => ((n+1)+a)/((n+1)+5) = (n+a)/(n+5) =>`  the sequence is constant

Hence, the monotony of sequence depends on a such that: the sequence decreases if `a > 5` , the sequence is constant if `a = 5`  and the sequence increases if `a < 5` .