Find the convergence domain and sum function of power series ∑ (∞, n → 0) (n + 1) x ^ n

Find the convergence domain and sum function of power series ∑ (∞, n → 0) (n + 1) x ^ n

First, the convergence radius r = LIM (n →∞) (n + 1) / (n + 2) = 1. Then, test X = 1, ∑ (n = 0, ∞) (n + 1) obvious divergence, test X = - 1, ∑ (n = 0, ∞) (- 1) ^ n * (n + 1) obvious divergence. Therefore, the convergence domain is (- 1,1). Let f (x) = ∑ (n = 0, ∞) (n + 1) * x ^ n be in (- 1,1). According to the term by term integral: ∫ (0, x) f (T) d

Finding the sum of power series (n = 1) NX ^ (n + 1) convergence domain and function

We can use the method of quadrature and derivation to find the sum function. The economic mathematics team will help you to solve it. Please evaluate it in time. Thank you!

The convergence domain of power series ∑ (∞ n = 1) (- 1) ^ n * [1] / [(3 ^ n) * (n)] * x ^ 2n + 1 is?

The absolute value of the limit of the latter term over the former = x ^ 2 / 3

1、 Find the convergence interval of the following power series 1. ∑ 10 ^ n × x ^ N 2. ∑ (- 1) ^ n × [x ^ (2n + 1)] / (2n + 1) 3. ∑ (X-5) ^ n / √ n
 4.∑1/(1+x^n) (x≠1)
2、 Find the sum function of the following series in the convergence interval
∑nx^(n-1)   
∑1/[(n+1)(n+2)]×x^(n+2)
X + x ^ 3 / 3 + X * 5 / 5 + ······ (- 1 & lt; X & lt; 1) and find the sum of series ∑ 1 / [(2n-1) 2 ^ n]

The convergence radius of power series ∑ x ^ n / N * 3 ^ n =?

Judging the convergence and divergence of ∑ sin (π / N) by comparison

The comparison criterion is used to compare with ∑ 1 / n

Convergence and divergence of series (∞Σ n = 1) 1 / N ^ n by comparison

The series is convergent. & nbsp; the economic mathematics team will help you solve it. Please evaluate it in time. Thank you!

Judging the convergence and divergence of series ∑ (n = 1, ∝) 2 ^ n sin (π / 3 ^ n)

∑(n=1,∝) 2^n sin(π/3^n)
When n tends to infinity, sin (π / 3 ^ n) ~ π / 3 ^ n
So the convergence and divergence of ∑ (n = 1, ∝) 2 ^ n sin (π / 3 ^ n) and ∑ (n = 1, ∝) 2 ^ n (π / 3 ^ n) = ∑ (n = 1, ∝) π (2 / 3) ^ n are the same
Because ∑ (n = 1, ∝) π (2 / 3) ^ n converges (3 π), the original order converges

Judging the convergence and divergence of series [∞Σ n = 1] sin [π / (2 ^ n)] by using the method of comparative convergence

Because when n tends to infinity, π / 2 ^ n tends to 0, so according to the substitution of equivalent infinitesimal: Sint &; t (T - > 0), sin [π / (2 ^ n)] & _; π / (2 ^ n) (n - > infinity), the convergence and divergence of [∞Σ n = 1] sin [π / (2 ^ n)] is the same as [∞Σn = 1] π / (2 ^ n), because 0 < 1

What is Leibniz's criterion of alternating series

The absolute value of the general term decreases and approaches zero