interval of convergence of $\sum n \exp (-x \sqrt n)$

Question

$$\sum^{\infty}_{n=1} n \exp (-x \sqrt n)$$

How to find the interval of convergence? Obviously, 0 is not in the interval because the series becomes divergent. could you help me?

Clearly, if it converges for $x_0$ then it converges on $(x_0,+\infty)$. — Thomas Andrews, Jul 05 '13 at 05:03
Because for larger $x$, each term is strictly smaller than for smaller $x$ — davidlowryduda, Jul 05 '13 at 05:06
'recreational-mathematics' seems entirely inappropriate for this question; that tag actually refers to a specific genre of math, not the motivations for the question. — Steven Stadnicki, Jul 09 '13 at 18:14

score 7 · Answer 1 · answered Jul 05 '13 at 05:06

7

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.

answered Jul 05 '13 at 05:06

André Nicolas

507,029

But this expansion only works when the exponent is small right? You are expanding about zero and $x \sqrt{n}$ can be arbitrarily large. Otherwise the comitted error is too big and you can not compare this expansion, am I right? – Jul 05 '13 at 06:38
1

It is late, but I don't think there is a problem. Yes, we are expanding $e^t$ about $0$, with $t=x\sqrt{n}$. And yes, for large $n$, truncating at the fifth term gives an awful approximation to $e^t$. But the point is that $e^t$ is bigger (after a while much bigger) than the estimate we made. That's good, it's in the denominator. That means that the convergence is in fact very fast, much much faster than the convergence of $\sum n^{-3/2}$. – André Nicolas Jul 05 '13 at 07:16
Yep, I realize that it is $e^{-x\sqrt{n}}$, you are right indeed. – Jul 09 '13 at 15:41

score 2 · Answer 2 · edited Apr 28 '16 at 15:43

2

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, $A= \lim_{n\to \infty} \frac {\log n}{\sqrt{n}} =0.$

edited Apr 28 '16 at 15:43

hardmath

37,015

answered Jul 05 '13 at 05:26

user64494

5,811

score 1 · Answer 3 · answered Jul 05 '13 at 05:20

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 $x$.

interval of convergence of $\sum n \exp (-x \sqrt n)$

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