How to prove that the series Σ f (n) {Q} / T (n) {P} has the same convergence as the series Σ 1 / N ^ (P-Q)? Where Q and P are the largest power of N in the function

How to prove that the series Σ f (n) {Q} / T (n) {P} has the same convergence as the series Σ 1 / N ^ (P-Q)? Where Q and P are the largest power of N in the function

F (n) {Q} / T (n) {P} is the quotient of two polynomials. The numerator is Q times and the denominator is p times. Now, the series Σ 1 / N ^ (P-Q) is used for comparison, so: Lim [f (n) {Q} / T (n) {P}] / [1 / N ^ (P-Q)] = Lim [f (n) {Q + P} / T (n) {P + Q}] = constant (i.e. the quotient of the coefficient of the highest power of the two polynomials), so the series Σ f (n) {Q} / T (n) {P

Continuous functions on open intervals must be bounded. A. error B. correct

A. Error, the limit at both ends of the open interval must exist before it is bounded

The function y = x / (x ^ 2 + 1) is bounded in (- ∞, + ∞). How to prove it

|y| = |x|/x^2+1
Y = 0 when x = 0
When x ≠ 0 | y | = 1 / (| x | + 1 / | x |)

How to prove that y = (x ^ 6 + x ^ 4 + x ^ 2) / (1 + x ^ 6) is a bounded function

When | x | ≤ 1, x ^ 6 ≤ 1, x ^ 4 ≤ 1, x ^ 2 ≤ 1, 0 ≤ y ≤ 3 / (1 + x ^ 6) ≤ 3
When | x | > 1, x ^ 4

It is proved that the function y = (x + 2) / (x ^ 2 + 1) is a bounded function Is there a way to avoid derivation.

1、 When x = - 2, y = 0. 2. When x ≠ - 2, let x + 2 = a, then: y = A / [(A-2) ^ 2 + 1] = A / (a ^ 2-4a + 5) = 1 / (a + 5 / A-4). 1. When a > 0, there is: Y > 0. Obviously, there are: a + 5 / a ≥ 2 √ 5, a + 5 / A-4 ≥ 2 √ 5-4, 11 / (a + 5 / A-4) ≤ 1 /

The derivative in the function, I only know a problem, and then find the derivative, find the slope, and the differential, I only know to find the derivative, and then add DX to the result. It feels too shallow. The book says that the derivative is to find the instantaneous quantity of independent variables and tangent problems, and I don't understand the differential at all,

Every function has a change trend, and the derivative is to find the change trend. Taking speed as an example, the derivative of speed to time is acceleration, that is, the speed of speed change