How to prove absolute and total convergence of $\sum_{n=1}^{+\infty}(-1)^n\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}$?

Question

Consider the series of functions $$\sum_{n=1}^{+\infty}(-1)^n\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}, \qquad x\in\mathbb{R}.$$

I want to study for which $x\in\mathbb{R}$ the series converges.

I start studying the absolute convergence. Taking $|f_n(x)| =\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}$, the necessary condition for the convergence is satisfied if $$\lim_{n\to +\infty}\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}\sim \lim_{n\to +\infty} \pi^{n(x-x^2)}=0 \iff x-x^2<0,$$ i.e. the series could converge if $x<0\cup x>1$.

${\bf EDIT:}$ In order to study the absolute convergence, I use the comparison. Thus, having that $\pi^{nx^2}+\sqrt{n}\ge \pi^{nx^2}$, we obtain that $$\sum_{n=1}^{+\infty}\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}\le \sum_{n=1}^{+\infty}\pi^{n(x-x^2)}$$ which does converge for $x<0\cup x>1$. Anyway, it is also true that $\pi^{nx^2}+\sqrt{n}\ge \sqrt{n}$ so that we could apply the comparison test in this way $$\sum_{n=1}^{+\infty}\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}\le \sum_{n=1}^{+\infty}\frac{\pi^{nx}}{\sqrt{n}}\sim\sum_{n=1}^{+\infty}\pi^{nx}$$ which does converge for $x<0$. As you see, the sets of convergence are different according to the different choices one make.

My question is: what is the correct answer? In which of the two set we obtain the absolute convergence?

Finally, I am in trouble also with total convergence.

${\bf 2nd EDIT}$: I start studying the total convergence when $x<0$. It is $$\sup_{x<0} \left\vert\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}\right\vert =\frac{1}{1+\sqrt{n}}$$ and the series $$\sum_{n=1}^{+\infty} \frac{1}{1+\sqrt{n}}$$ diverges, so we have no total convergence in $(-\infty, 0)$. Thus, I proceed studying the convergence in the compact subsets $(1, a]\subset (1, +\infty)$. It is

$$\sup_{x\in (1, a]} \left\vert\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}\right\vert= \frac{\pi^{na}}{\pi^{na^2}+\sqrt{n}}$$

and the series $$\sum_{n=1}^{+\infty}\frac{\pi^{na}}{\pi^{na^2}+\sqrt{n}}$$

does converge. Could anyone please tell me if my resoning is correct?

Thank you in advance!

Answer 1

It seems that you mapped out two alternatives to dominate your original series with a convergent one.

1. The absolute value of the denominator of the original series is larger than one for $n\geq1$. Then, you may use $\sum_n\pi^{nx}$ as a comparator, you get a geometric series ($z=\pi^x$) which can be seen to converge for $x<0$ (uniform convergent in compact subsets of $(-\infty,0)$

2. The absolute of the denominator of the original series is larger than $\pi^{nx^2}$ you use $\sum_n\pi^{n(x-x^2)}$ as a comparator, this is again a geometric series ($z=\pi^{x-x^2}$) which converges if $\pi^{x-x^2} <1$. Thus, you get convergence for $x\in (-\infty,0)\cup(1,\infty)=U$ (uniform convergence in compact subsets of $U$).

The second option gives you a larger domain of convergence.

Alternatively, since the expressions are not very complicated, you may also try to use a convergence test, such as the ratio test, on the original series. Since $\pi^{x^2}>1$ for all $x\neq0$ $$\frac{\frac{\pi^{(n+1)x}}{\pi^{(n+1)x^2}+\sqrt{n+1}}}{\frac{\pi^{nx}}{\pi^{nx^2}+\sqrt{n}}}=\frac{\pi^{nx^2}+\sqrt{n}}{\pi^{(n+1)x^2}+\sqrt{n+1}}\pi^x\xrightarrow{n\rightarrow\infty}\pi^{x-x^2}.\qquad x\neq0$$ which gives you converges for all $x$ fir which $\pi^{x-x^2}<1$ (same as in (2)).