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Advanced Mathematics, Existence of Limit of Sequence, Method 4: Monotone Method, Why does a 1≥a2 still get f'(x)≥0?
The correctness of this conclusion is based on the monotonicity principle of functions.

If the derivative f'(x) of the function f(x) is greater than or equal to 0, then f(x) monotonically increases at x.

Therefore, if the series {a(n)} has the recurrence relation a(n+ 1)=f(a(n)) and the derivative f'(x) of the function f(x) is greater than or equal to 0, the series {a(n)} must increase monotonically.

Similarly, if the derivative f'(x) of the function f(x) is less than or equal to 0, then f(x) monotonically decreases at x.

Therefore, if the series {a(n)} has the recurrence relation a(n+ 1)=f(a(n)) and the derivative f'(x) of the function f(x) is less than or equal to 0, the series {a(n)} must be monotonically decreasing.

For the sequence {a(n)}, each item a(n) in it

) is the value of the independent variable x of the function f(x). Therefore, when f'(x) is greater than or equal to 0, for any x 1 and x2 (where x 1

So, if the series {a(n)} is a decreasing series.

(a( 1)>=a(2)), even if f'(x) is still >; =0, the sequence {a(n)} will not increase monotonically, but decrease monotonically.

This is because the monotonicity of the sequence {a(n)} depends on the monotonicity of the recurrence relation a(n+ 1)=f(a(n)) and the function f(x), but not on the initial value of the sequence {a(n)}. Therefore, even if the initial value of the sequence {a(n)} is decreasing, as long as the derivative f'(x) of the function f(x) is less than or equal to 0, the sequence {a(n)} will still decrease monotonically.

Therefore, the conclusion is correct.