# convergence of $\sum_{n=1}^{\infty} \frac{e^{-xn\ln x}}{x^2 + n}$ if $x>0$

i'm supposed to study the following series: $$\sum_{n=1}^{\infty} \frac{e^{-xnlnx}}{x^2 + n}$$ as $$x>0$$

so I started by rewriting the series as $$\sum_{n=1}^{\infty}$$ $$\frac{1}{(e^{xnlnx})(x^2 + n)}$$ by eliminating the minus at numerator

then by logarithm properties, since $$e^{xnlnx}$$=$$(e^n)(e^{ln(x^x)})$$=$$(e^n)(x^x)$$

we can rewrite the series as: $$\sum_{n=1}^{\infty} \frac{1}{(e^n)(x^x)(x^2 +n)}$$

notice $$x^x>0$$ $$\forall x>0$$ , not just that but also notice that by computing his derivative it results decresing for $$0 and increasing for $$x>1/e$$. thus it has a global minimum in $$1/e$$ in our domain. thus it follows that:

$$\frac{1}{(e^n)(x^x)(x^2 +n)}$$ $$\leq$$ $$\frac{1}{(e^n)(1/e)(x^2 +n)}$$ and this holds for all $$x>0$$

so by comparison test it suffices to show that :

$$\sum_{n=1}^{\infty} \frac{1}{(e^n)(1/e)(x^2 +n)}$$ is covergent to show that our series is covergent, indeed by rewriting it as:

$$e \sum_{n=1}^{\infty} \frac{1}{(e^n)(x^2 +n)}$$ since we have that $$(x^2 +n)>1$$ by hypothesis then our series is again by comparison test convergent if the series $$\sum_{n=1}^{\infty} \frac{1}{(e^n)}$$ is.

notice this is obvioulsy convergent by geometric series test, so we conclude that our series converges $$\forall x>0$$

is my reasoning sound or I'm missing somenthing? thanks in advance

• Something is wrong : set $x=1$ in the series and check? Nov 28 '21 at 9:38
• I just want to note that you could've just factored the $x^{-x}$ term out, since it wasn't dependent on $n$. There was no need to bound that, it doesn't affect the convergence of the series. Nov 28 '21 at 9:39
• @TeresaLisbon you are right Nov 28 '21 at 9:44
• I think the logarithm rule is used wrongly here. You should get $e^{-xn\ln x} = (e^{\ln x})^{-nx} = x^{-nx}$. Basically, you have $a^{bc} = (a^{b})^c$ and not $a^ba^c$. Nov 28 '21 at 9:49
• @invictus Welcome! Although that still means we haven't solved the problem. Nov 28 '21 at 10:29

By substitution $$a=x^{-x}$$ for $$x>1$$ you obtain a series $$\sum_{n=1}^{\infty}\frac{a^n}{b+n},~00$$. Using the ratio test you immediately obtain convergence: $$\left|\frac{\frac{a^{n+1}}{b+n+1}}{\frac{a^n}{b+1}}\right|=a\left|\frac{b+n}{b+n+1}\right|<1$$. For $$x=1$$ you obtain a harmonic series, which is known to diverge. And finally for $$0, for which $$a=x^{-x}>1$$, by using the ratio test again you can find sufficiently high $$n$$ that the ratio is greater than $$1$$, ergo the given series diverges as well. So the series converges for $$x>1$$, otherwise diverges.
I appear to get a different answer. $$a_n = \frac{x^{-nx}}{x^2+n}$$ $$\implies \lim_\limits{n \to \infty} (a_n)^{\frac{1}{n}} = \lim_\limits{n \to \infty} \frac{x^{-x}}{(x^2+n)^\frac{1}{n}}$$ $$= \lim_\limits{n \to \infty} \frac{x^{-x}}{x^{\frac{2}{n}}(1+\frac{n}{x^2})^\frac{1}{n}}$$ Now $$x^{\frac{2}{n}} \to 1$$ and $$\Big(1+\frac{n}{x^2}\Big)^\frac{1}{n} \to e^{\frac{n}{nx^2}} = e^{\frac{1}{x^2}}$$
$$\implies \lim_\limits{n \to \infty} (a_n)^{\frac{1}{n}} = x^{-x} e^{\frac{1}{x^2}}$$
Plugging this in a graphing calculator (or equivalently analyzing using derivatives) shows that its always lesser than 1. So, shouldn't the series converge $$\forall x \in \mathbb{R}^+$$