$\sum_\limits{n=1}^{\infty}\frac{n}{x^n}$ ratio or root test? different results?

Question

Show that $\sum_\limits{n=1}^{\infty}\frac{n}{x^n}$ converges absolutely.

I used the ratio test:

$\lim_{n\to\infty}\frac{\frac{n}{x^n}}{\frac{n+1}{x^{n+1}}}=\lim_{n\to\infty}\frac{n}{n+1}x=x$, so the dominion where the function converges is $|x|<1$.

However the resolution proposed to use the root test:

$\lim_{n\to\infty}(\frac{n}{|x|^n})^{\frac{1}{n}}\leqslant\lim_{n\to\infty}(\frac{n}{|x|^n})^{\frac{1}{n}}=\frac{1}{|x|}$ therefore the function converges when $\frac{1}{|x|}>1\implies |x|>1$ which contradicts the ratio test.

Question:

Why are these two methods delivering different results regarding the convergence dominion? Which one is wrong? Why is that one wrong?

Thanks in advance!

Answer 1

You have calculated $\large \lim {a_n\over a_{n+1}}$ instead of $\large \lim {a_{n+1}\over a_n}$ therefore the result must be $\ge 1$. There is no contradiction.