Furthermore,1= lim {n→∞}1-(2/3) n ≤ lim {n→∞} (1+(-2/fu3) n) (1/n) ≤ lim {n)
So lim {n →∞} (1+(-2/3) n) (1/n) =1.
Lim {n →∞} n ( 1/n) = 1。
get lim { n→∞}((3n+(-2)n)/n)( 1/n)= 3？ lim { n→∞}(( 1+(-2/3)^n)/n)^( 1/n)= 3。
The convergence radius of power series is 1/3.
Only the convergence and divergence at the end points are discussed.
For x = 1/3, the general term is (1+(-2/3) n)/n, which is a positive sequence equivalent to 1/n, and its divergence can be known by comparison and discrimination.
For x =-1/3, the general term is (-1) n+(2/3) n)/n. ∑ (2/3) n/n and ∑ (- 1) n/n converge, so x.
The convergence domain of the synthesis is [- 1/3, 1/3].
Question 2: The main process of finding the convergence interval of power series in a graph is to use D 'Alembert discriminant method, and the power series is absolutely convergent within the convergence radius.
1. Item n: (x-6) n/(n! ? 6^n).
Item n+1:(x-6) (n+1)/((n+1)! ? 6^(n+ 1)).
When n → ∞, the ratio (x-6)/(6(n+ 1)) → 0 holds for any X. 。
So the convergence interval of power series is (-∞,+∞).
2. item n: n! (x-6)^n/6^n.
Item n+ 1: (n+ 1)! (x-6)^(n+ 1)/6^(n+ 1).
When n → ∞, the ratio (n+ 1)(x-6)/6 → ∞ holds for any x ≠ 6.
So the power series converges only when x = 6.
3. Item N: (x-6) n/6 n
Item n+1:(x-6) (n+1)/6 (n+1). The ratio is constant (x-6)/6.
When |x-6| 6, the series diverges.
When |x-6| = 6, the general term of the series does not converge to 0, so the series diverges.
So power series is only for |x-6| Question 3: What is the difference between the convergence domain and the convergence interval of power series? How to find the convergence interval is an open interval, and the convergence domain is to judge whether it converges at the end of the convergence interval.
Just as you find that the convergence radius of a series is 5, then the convergence interval at this time is (-5,5). The next step is to find the convergence region of x =-5 and x = 5, depending on whether it converges.
For example, when x =-5 converges and x = 5 diverges, the convergence domain is [-5,5].
Question 4: How to find the convergence interval 9? Divided into two power series.
Find the convergence radius respectively.
If the radius is small, calculate the convergence interval.
The process is as follows:
Question 5: To find the sum function of power series, the convergence interval of power series is 100. Hello! The sum function can be obtained by the derivative quadrature method as shown in the figure, and the convergence region needs to be discussed first. The economic mathematics team will help you solve the problem, please adopt it in time. thank you
Question 6: What is the difference between the convergence domain and the convergence interval of power series? Assuming that the convergence radius r of power series has been found,
The convergence interval of power series is an open interval (-R, r);
Then it is judged whether the power series converges to x= -R and x=R,
Considering these two points, that is, the two endpoints (-R, r) of the open interval, it is the convergence domain.
For example, if it converges at x= -R and diverges at x=R, the convergence domain is [-R, R].