$\sum_\limits{n=1}^{\infty}\frac{(2n-1)^{n}(x+1)^n}{2^{n-1}n^n}$ converegence domain

Question

Exercise:Find the convergence domain of the series $$\sum_\limits{n=1}^{\infty}\frac{(2n-1)^{n}(x+1)^n}{2^{n-1}n^n}$$

I tried to solve it in the following way:

Applying the comparison test:

$$\sum_{n=1}^{\infty}\frac{(2n-1)^{n}(x+1)^n}{2^{n-1}n^n}\leqslant\sum_{n=1}^{\infty}\frac{(2n-1)^{n}(x+1)^n}{n^n}$$

Now applying the root test:

$$\lim_{n\to\infty}\frac{(2n-1)^{}(x+1)}{n}=2(x+1)$$

$$2x+1<1\implies x<-\frac{1}{2}$$

Solution(Book):$-2<x<0$

Question:

What am I doing wrong? What should I do?

Answer 1

Apply the ratio test directly:$$\sqrt[n]{\frac{(2n-1)^n|x+1|^n)}{2^{n-1}n^n}}=\sqrt[n]2\frac{(2n-1)|x+1|}{2n}\to|x+1|.$$Therefore, the series converges if $|x+1|<1$ and it diverges if $|x+1|>1$.

If $|x+1|=1$, then$$\left|\frac{(2n-1)^n|x+1|^n)}{2^{n-1}n^n}\right|=2\left(1-\frac1{2n}\right)^n\to\frac2{\sqrt e}\neq0.$$Therefore the series diverges if $|x+1|=1$. So, the domain of convergence is indeed $(-2,0)$.

Answer 2

Your problem is incorrectly applying the root test. You get $$ L = \lim_{n \to \infty} a_n^{1/n} = \lim_{n \to \infty} \frac{(2n-1)(x+1)}{2n} \frac{1}{2^{1/n}} = x+1, $$ and you must force $|L|<1$ which yields $|x+1|<1$ with solution $-2<x<0$.