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!

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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
$$x^{\sqrt{n}}=\frac{1}{e^{y\sqrt{n}}}\lt \frac{1}{e^{ky\log n}}=\frac{1}{n^{ky}}\lt \frac{1}{n^2}.$$

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.

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 $x \in (0,1)$.

As Adam mentioned, the integral test will work. Just have to be careful that everything being done with respect to $n$ and remember that $0<x<1$. Let’s fix $x$ between zero and one and we see that $f(n)=x^{\sqrt{n}}$ is non-negative and monotonically decreasing so we can use the integral test.

We need $$\int_1^{\infty} x^{\sqrt{n}} dn = \int_1^{\infty} e^{\sqrt{n} \ln(x)}dn.$$

Remember that $\ln(x)$ is a negative constant here so the integral does converge. But if you want to go ahead and explicitly show it, you can do it. Just use the substitution
$$u = \sqrt{n}$$
$$ du = \frac{dn}{2\sqrt{n}}$$
$$2udu=dn$$

and then integrate by parts and undo the substitution and you get

$$\left.\frac{2x^{\sqrt{n}}(\sqrt{n} \ln(x)-1)}{(\ln(x))^2}\right]_{n=1}^{\infty}$$

which goes to zero as $n\rightarrow\infty$ (because $0<x<1$) and plugging in $n=1$ simplifies it so you get
$$0-\frac{2x(\ln(x)-1)}{(\ln(x))^2}$$ which of course is finite for our fixed $x$. Therefore the original series converges because this integral converges.

Try the integral test ! $\quad\displaystyle\sum_{n=1}^\infty a^{-\sqrt[k]n}~=~-1+\sum_{n=0}^\infty a^{-\sqrt[k]n}\quad{\large\approx}~-1+\int_0^\infty a^{-\sqrt[k]t}~dt~=~\frac{k!}{\ln^ka}-1.$

See $\Gamma$ function for more details. In this case, $a=\dfrac1x$ and $k=2$.