How to find the sum function of (n = 1, ∞) Σ n * x ^ (n-1) power series? I think the answer is 1 / (1-x). Why is 2n missing?

How to find the sum function of (n = 1, ∞) Σ n * x ^ (n-1) power series? I think the answer is 1 / (1-x). Why is 2n missing?

The sum function is a function of X, independent of n

Sum function of power series ∑ (∞ n = 1) (x ^ n) / n

First, the series ∑ (∞ n = 1) (x ^ n) / N is derived term by term, and D (∑ (∞ n = 1) (x ^ n) / N) DX = ∑ (∞ n = 0) x ^ n is obtained. When | x | 1, the series converges, and the sum function s (x) = 1 / (1-x), that is, D (∑ (∞ n = 1) (x ^ n) / N) DX = s (x) = 1 / (1-x), and the product of both ends is ∑ (∞ n = 1) (x ^ n) / N = - ln (1-x) + C (C is constant), Then we substitute x = 0 into the formula to get C = 0, so we get the conclusion ∑ (∞ n = 1) (x ^ n) / N = - ln (1-x)

How to find the sum function of power series X ^ n / N! 2 ^ n, from n = 0 to n = ∞

　　 Σ(n=0~∞)x^n/n!2^n
　　= Σ(n=0~∞)(x/2)^n/n! = e^(x/2), -inf.