$\sum_\limits{n=1}^{\infty}\frac{1}{n!x^n}$ and $\sum_\limits{n=1}^{\infty}\frac{1}{(2n-1)x^n}$ different convergence domains?

Question

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

First series: $\sum_\limits{n=1}^{\infty}\frac{1}{n!x^n}\leqslant\sum_\limits{n=1}^{\infty}\frac{1}{x^n}$ and now employing the Cauchy integral test:

$\int_\limits{1}^{\infty}\frac{1}{x^n}dn=\frac{1}{x\log(x)}$

So the series converge absolutely for $x\neq 0$.

Solution(book):"series converge absolutely for $x\neq 0$"

Second series:

$\sum_\limits{n=1}^{\infty}\frac{1}{(2n-1)x^n}\leqslant\sum_\limits{n=1}^{\infty}\frac{1}{x^n}$ since $2n-1>1$. So the series should converge absoulutely on $x\neq 0$.

Solution(book): $x>1\:\:,\:\:x\leqslant -1$.

Question:

Why does the solutions different? Am I doing something wrong?

Answer 1

Your approach to first series is wrong. Simply apply the quotient test and you'll see that it converges always. Your approach could not possibly work because $\sum_{n=1}^\infty\frac1{x^n}$ only converges when $|x|>1$.

Applying the quotiente test to the second series, you'll see that it converges if $|x|>1$ and that diverges if $|x|<1$. Finally, if $x=1$ it diverges (like the harmonic series) and if $x=-1$ it converges, by the Leibniz test.