Let f (x) be differentiable in the interval (0,1) and f '(x) be bounded. It is proved that the series ∑ (n from 2 to infinity) [f (1 / N) - f (1 / (n + 1))] is absolutely convergent In the answer, [f (1 / N) - f (1 / (n + 1)) = f '(ζ) (1 / n-1 / (n + 1)) = f' (ζ) * 1 / N (n + 1),) absolute value f (1 / N) - f (1 / (n + 1)) ≤ M / N ^ 2, how does this m / N ^ 2 come from

Let f (x) be differentiable in the interval (0,1) and f '(x) be bounded. It is proved that the series ∑ (n from 2 to infinity) [f (1 / N) - f (1 / (n + 1))] is absolutely convergent In the answer, [f (1 / N) - f (1 / (n + 1)) = f '(ζ) (1 / n-1 / (n + 1)) = f' (ζ) * 1 / N (n + 1),) absolute value f (1 / N) - f (1 / (n + 1)) ≤ M / N ^ 2, how does this m / N ^ 2 come from

The second inequality is the intermediate value theorem of continuous function. The maximum value of F '(ζ) less than f' (x) is M. and 1 / N (n + 1) less than 1 / N ^ 2. Because 1 / N ^ 2 converges, so 1 / N (n + 1) converges. So the absolute value f (1 / N) - f (1 / (n + 1)) converges. Then [f (1 / N) - f (1 / (n + 1))] absolutely converges

Let f (x), G (x) be bounded in the domain D. try to prove that f (x) ± g (x), f (x) · g (x) are also bounded in the domain D

It is known that there exists m > 0 such that for any x ∈ D, | f (x) | ≤ m, | g (x) | ≤ M
So | f (x) ± g (x) | ≤ | f (x) | + | g (x) | = 2m
|f(x)g(x)|=|f(x)||g(x)|≤M²
So we can see that f (x) ± g (x), f (x) · g (x) is bounded on D!

If the functions f and G are bounded on the domain D, it is proved that F + G, F-G and FG are also bounded on D

Let | F|