interval of convergence of $\sum n \exp (-x \sqrt n)$ $$\sum^{\infty}_{n=1} n \exp (-x \sqrt n)$$
How to find the interval of convergence? Obviously, 0 is not in the interval because the series becomes divergent. could you help me?
 A: If $x$ is negative, there is obvious divergence.
For positive $x$, the general term is $\dfrac{n}{e^{x\sqrt{n}}}$. Using the power series expansion of the exponential, we can see that
$$e^{x\sqrt{n}}\gt 1+n^{1/2}x +\frac{nx^2}{2!} +\cdots +\frac{n^{5/2}x^5}{5!} \gt \frac{n^{5/2}x^5}{5!}.$$
It follows that the $n$-th term of our sequence is less than $\dfrac{5!}{x^5} \dfrac{1}{n^{3/2}}$.
But we know that $\displaystyle\sum \frac{1}{n^{3/2}}$ converges. Thus, by Comparison, so does our series.
A: This is a Dirichlet series. Up to Encyclopedia of Mathematics , the abscissa of its absolute convergence $$A=\limsup_{n\to\infty} \frac {\log |a_n|} {\lambda_n}. $$ In your case $a_n=n$ and $\lambda_n=\sqrt{n}.$ Thus,
$A= \lim_{n\to \infty} \frac {\log n}{\sqrt{n}} =0.$
A: Lemma: If $p(n)$ is a non-zero polynomial, then $\sum p(n)z^n$ has a radius of convergence $1$.
I'll assume we know this.
Now if $x>0$, then: $$\sum_{n=k^2+1}^{(k+1)^2} ne^{-x\sqrt{n}} \leq \sum_{n=k^2+1}^{(k+1)^2} (k+1)^2 e^{-kx}= (k+1)^2(2k+1)e^{-kx}$$
So this means that $$\sum_{n=1}^{(m+1)^2} ne^{-x\sqrt{n}} \leq \sum_{k=0}^{m+1} (k+1)^2(2k+1)e^{-kx}$$
Letting $z=e^{-x}$ and $p(n)=(n+1)^2(2n+1)$ then this is $\sum_{k=0}^{m+1} p(k)z^k$. If $x>0$ then $|z|<1$.
So $\sum_{k=0}^\infty p(k)z^k$ converges, and all the terms of the original series are positive, then $\sum_{n=1}^\infty ne^{-x\sqrt{n}}$ converges, too.
So it converges for any $x>0$ and only those $x$.
