convergence of a series involving $x^\sqrt{n}$

I was trying to prove the convergence of the series $\sum_{n=1}^{\infty}x^{\sqrt{n}}$, for $0<x<1$. Unfortunately, I could not make one of the standard convergence tests give me an answer. Does anybody of you have a suggestion? any help is much appreciated!

many thanks!

Solution 1:

Let $x=\dfrac{1}{e^y}$. Note that $y$ is positive. There is an integer $k$ such that $ky\gt 2$.

For large enough $n$, $\sqrt{n}\gt k\log n$. We can if we wish prove this using L'Hospital's Rule, by showing that $\lim_{n\to\infty} \frac{\log n}{\sqrt{n}}=0$;

For any such $n$. we have $$x^{\sqrt{n}}=\frac{1}{e^{y\sqrt{n}}}\lt \frac{1}{e^{ky\log n}}=\frac{1}{n^{ky}}\lt \frac{1}{n^2}.$$

Solution 2:

Let us compute $$ n^2x^\sqrt{n}=e^{2\ln n+\ln x\sqrt{n}}=e^{\ln x \sqrt{n}\left(1+\frac{2\ln n}{\ln x\sqrt{n}}\right)}. $$ Thus we see that $$ \lim_{n\rightarrow +\infty} n^2x^\sqrt{n}=0. $$ So the sequence $(n^2x^\sqrt{n})$ is bounded and there exists $C>0$ such that $$ 0\leq x^\sqrt{n}\leq \frac{C}{n^2} $$ for all $n\geq 1$. Then the convergence of $\sum_{n\geq 1} x^\sqrt{n}$ follows by comparison.

Solution 3:

For $x \in (0,1)$, we have $$x^{\sqrt{n}} \leq x^k \,\,\,\,\,\,\,\,\,\,\ \forall n \geq k^2$$ $$\sum_{n=k^2}^{(k+1)^2-1}x^{\sqrt{n}} \leq 2kx^k$$ Hence, we have $$\sum_{n=1}^{(N+1)^2-1} x^{\sqrt{n}} \leq \sum_{n=1}^N 2nx^n$$ Hence, $$\sum_{n=1}^{\infty} x^{\sqrt{n}} $$ converges for $x \in (0,1)$ since $ \displaystyle \sum_{n=1}^\infty 2nx^n$ converges for $x \in (0,1)$.