Convergence of series $\frac{x^{n}}{\sqrt{n}}$ when $0<x<1$

Question

I was stuying 1-D random walk example in Borel-Cantelli Lemma

Event $A_n$ = We return back to origin in 2n steps

Probability of taking right step at each point = p

Probability of taking left step at each point = q

$\sum_{n=1}^{\infty}(P(An)) =$ $2n\choose n$ $(p^n)(q^n)$

After using sterling's approximation it comes to the form: $k\cdot \sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$

So I need to find if series $\sum_{n=1}^{\infty} \frac{x^{n}}{\sqrt{n}}$ when $0 < x < 1$ converges or not to determine which Borel-Cantelli Lemma would be applicable.

Answer 1

$\sum \frac {x^{n}} {\sqrt n} \leq \sum x^{n} =\frac x {1-x} <\infty$.

Answer 2

The series converges for $-1<x<1$. This can be established with the Ratio Test.

\begin{align*} \lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{\sqrt{n+1}}}{\frac{x^n}{\sqrt{n}}}\right| &= \lim_{n\to\infty}\left|\frac{\sqrt{n}}{\sqrt{n+1}}x\right|\\ &= |x|\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}\\ &= |x|<1\text{ if } -1<x<1 \end{align*}