What is convergence interval of $ \sum_{n=1}^\infty a^{\sqrt{n}} \cdot x^n$ where $a \in (0,1)?$

Question

What is convergence interval of $ \sum_{n=1}^\infty a^{\sqrt{n}} \cdot x^n$ where $a \in (0,1)?$

So far, I´ve got radius $$ R^{-1} = \lim_{n\to\infty} \left|a^{\sqrt{n+1}-\sqrt{n}}\right| = 1.$$

Thus interval of convergence is $(-1,1)$. But now, I need to investigate convergence of $$ \sum_{n=1}^\infty a^{\sqrt{n}} $$ and $$ \sum_{n=1}^\infty a^{\sqrt{n}} \cdot (-1)^n. $$ I can't see, what criterion I should use.

Answer 1

Two hints:

1. If a series converges, then its terms converge to zero.

2. If a positive decreasing sequence converges to zero, then the sum of its terms taken with alternating signs converges.

Also, keywords: comparison test.

edit

Ok, seems, the hint about comparison test was a bit inevident. We will assume that $a\in (0,1).$

Let us show that there exists a positive constant $C>0$ such that $\forall n\ge 1$ we have $$a^{\sqrt{n}}\le \frac{C}{n^2}.$$ It is sufficient to show that $$\forall n\ge 1 \quad \ln C -2\ln n - \sqrt{n}\ln a\ge 0.$$ We know that $\lim _{n\to \infty}(-\sqrt{n}\ln a-2\ln n)=+\infty$ (because $a\in(0,1)$), therefore we can take $$\ln C=\inf_{n\in\Bbb N}(-\sqrt{n}\ln a-2\ln n)>-\infty.$$

Armed with this inequality, we can now say that $$\sum_{n\ge 1}a^{\sqrt{n}}\le \sum_{n\ge 1}\frac{C}{n^2}\le \frac{\pi^2C}{6}.$$

edit 2

To reiterate upon the comparison and geometric series.

We know that $$\sum_{k^2\le n<(k+1)^2}a^{\sqrt{n}}\le ((k+1)^2-k^2)a^{k},$$ therefore $$\sum_{n\ge 1}a^{\sqrt{n}}\le \sum_{k\ge 1} ((k+1)^2-k^2)a^{k} = \sum_{k\ge 1} ( 2k+1)a^{k}, $$and the latter geometric series converges.

Answer 2

The integral test works. Let $b=\ln a$. Then $b<0$ and $a^{\sqrt{x}}=e^{b\sqrt{x}}$ is continuous, positive and decreasing. So, we compute $$I=\int_1^{\infty}e^{b\sqrt{x}}dx$$ Let $u=\sqrt{x}$, so $du=dx/2\sqrt{x}=dx/2u$. When $x=1$, $u=1$ and as $x\to\infty$, $u\to\infty$. Therefore, \begin{align*} I&=2\int_1^{\infty}ue^{bu}du=\lim_{N\to\infty}{2}(u/b-1/b^2)e^{bu}|_1^{N} =\lim_{N\to\infty}2(N/b-1/b^2)e^{bN}-2(b-1)e^b/b^2.\\ \end{align*} Since $b<0$, $\lim_{N\to\infty}(N/b-1/b^2)e^{bN}=0$, and $I=2(1-b)e^b/b^2$. Therefore, by the integral test, $$\sum_{n=1}^\infty a^{\sqrt{n}}$$ converges.

It follows that $\sum (-1)^na^{\sqrt{n}}$ is absolutely convergent, so it is convergent as well.