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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 $0<x<1/e$ 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

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    $\begingroup$ Something is wrong : set $x=1$ in the series and check? $\endgroup$ Nov 28 '21 at 9:38
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    $\begingroup$ 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. $\endgroup$ Nov 28 '21 at 9:39
  • $\begingroup$ @TeresaLisbon you are right $\endgroup$
    – invictus
    Nov 28 '21 at 9:44
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    $\begingroup$ 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 $a^ba^c$. $\endgroup$ Nov 28 '21 at 9:49
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    $\begingroup$ @invictus Welcome! Although that still means we haven't solved the problem. $\endgroup$ Nov 28 '21 at 10:29
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By substitution $a=x^{-x}$ for $x>1$ you obtain a series $\sum_{n=1}^{\infty}\frac{a^n}{b+n},~0<a<1,~b>0$. 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 $0<x<1$, 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.

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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 $\forall x \in \mathbb{R}^+$

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  • $\begingroup$ Indeed your answer is correct, as Teresa pointed out in the comments I made a misuse of the logarithm property at the beginning. $\endgroup$
    – invictus
    Nov 28 '21 at 12:22

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