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Let $\varepsilon \in (0, 1)$ and consider the analytic function $$f(x) = \sum_{n=1}^\infty n^\varepsilon \frac{x^n}{n!}.$$ What is the order of growth of $f(x)$ as $x \to \infty$? From the basic inequality $1 \leqslant n^\varepsilon \leqslant n$ I deduced $$e^x - 1 < f(x) < xe^x$$ but I would like an asymptotic equivalence $$f(x) \sim g(x) e^x \qquad (x \to \infty)$$ where $g(x)$ is sufficiently familiar (e.g. a combination of $\log$s). How do I proceed?

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  • $\begingroup$ Wolfram Alpha doesn't even have an answer for $x=1$ ! $\endgroup$
    – user65203
    Jul 31 '19 at 15:26
  • $\begingroup$ $g(x)$ will depend on $\epsilon$. $\endgroup$ Jul 31 '19 at 16:31
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    $\begingroup$ We have the better asymptotic equivalence$$f(x)\sim x^{\varepsilon}e^x\qquad(x\to\infty)$$ $\endgroup$ Jul 31 '19 at 16:39
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    $\begingroup$ In one of his blog posts, Terry Tao says that it's not a bad rule of thumb to look at the maximal terms of series for their asymptotics. Since the largest term in your case is roughly at n=x, your series should be roughly $x^\epsilon$ times as large as $e^x$. I hope I'm not misquoting here, would love to see that blog post again and see if it says more that is applicable to your situation $\endgroup$
    – Bananach
    Jul 31 '19 at 17:43
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    $\begingroup$ Also, Sangchul Lee's answer here can be easily adapted to your case. $\endgroup$ Jul 31 '19 at 18:30
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Let $\{a_n\}_{n=0}^{\infty}$ and $\{b_n\}_{n=0}^{\infty}$ be sequences of real numbers such that:

  • $b_n$ is eventually positive (i.e. $b_n>0$ for all $n>n_0$);

  • Both $a(x)=\sum\limits_{n=0}^{\infty}a_n x^n$ and $b(x)=\sum\limits_{n=0}^{\infty}b_n x^n$ converge for all $x$.

Then $\lim\limits_{n\to\infty}a_n/b_n=\lambda$ implies $\lim\limits_{x\to+\infty}a(x)/b(x)=\lambda$.

For a proof, we can assume $\lambda=0$ (otherwise replace $a_n$ with $a_n-\lambda b_n$). Now let $\varepsilon>0$, and $|a_n/b_n|<\varepsilon$ when $n>N$. Increasing $N$ if needed, we can assume $b_n>0$ when $n>N$. Then $$|a(x)|\leqslant\left|\sum_{n=0}^{N}a_n x^n\right|+\varepsilon\left(b(x)-\sum_{n=0}^{N}b_n x^n\right),$$ i.e. $\limsup\limits_{x\to+\infty}|a(x)/b(x)|\leqslant\varepsilon$. As $\varepsilon>0$ is arbitrary, we're done.

For $a>0$ we have $\Gamma(a)=\lim\limits_{n\to\infty}n^a\mathrm{B}(n, a)$.

This can be shown by taking $\lim\limits_{n\to\infty}$ of $$n^a\mathrm{B}(n,a)=n^a\int_0^1 t^{a-1}(1-t)^{n-1}\,dt=\int_0^n x^{a-1}(1-x/n)^{n-1}\,dx.$$


Now write $f(x)=x\sum_{n=0}^{\infty}(n+1)^{\varepsilon-1}x^n/n!=xa(x)$ with $$a_n=\frac{(n+1)^{\varepsilon-1}}{n!},\qquad b_n=\frac{\mathrm{B}(n+1,1-\varepsilon)}{n!\ \Gamma(1-\varepsilon)}.$$ We have $\lim\limits_{n\to\infty}a_n/b_n=1$ and then $\lim\limits_{x\to+\infty}f(x)/(xb(x))=1$. But \begin{align}xb(x) &=\frac{x}{\Gamma(1-\varepsilon)}\sum_{n=0}^{\infty}\frac{x^n}{n!}\int_0^1 t^n(1-t)^{-\varepsilon}\,dt \\&=\frac{x}{\Gamma(1-\varepsilon)}\int_0^1(1-t)^{-\varepsilon}e^{xt}\,dt \\&=\frac{x^{\varepsilon}e^x}{\Gamma(1-\varepsilon)}\int_0^x z^{-\varepsilon}e^{-z}\,dz \end{align} (after substituting $t=1-z/x$). Thus $\color{blue}{\lim\limits_{x\to+\infty}f(x)/(x^{\varepsilon}e^x)=1}$.


Yet another approach, allowing better asymptotics: \begin{align} f(x)&=x\sum_{n=0}^{\infty}\frac{x^n}{n!}(n+1)^{\varepsilon-1} \\&=\frac{x}{\Gamma(1-\varepsilon)}\sum_{n=0}^{\infty}\frac{x^n}{n!}\int_0^1 t^n(-\ln t)^{-\varepsilon}dt \\&=\frac{x}{\Gamma(1-\varepsilon)}\int_0^1 e^{xt}(-\ln t)^{-\varepsilon}dt \\&=\frac{xe^x}{\Gamma(1-\varepsilon)}\int_0^1 e^{-xt}\big(-\ln(1-t)\big)^{-\varepsilon}dt \end{align} and now we expand $\big(-\ln(1-t)/t\big)^{-\varepsilon}$ into power series: $$\big(-\ln(1-t)/t\big)^{-\varepsilon}=1-\frac{\varepsilon}{2}t-\frac{5\varepsilon-3\varepsilon^2}{24}t^2-\frac{6\varepsilon-5\varepsilon^2+\varepsilon^3}{48}t^3\pm\ldots$$ which, by Watson's lemma, gives $$f(x)\asymp x^{\varepsilon}e^x\left(1-\frac{\varepsilon(1-\varepsilon)}{2x}-\frac{\varepsilon(1-\varepsilon)(2-\varepsilon)(5-3\varepsilon)}{24x^2}-\frac{\varepsilon(1-\varepsilon)(2-\varepsilon)^2(3-\varepsilon)^2}{48x^3}\pm\ldots\right)$$

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    $\begingroup$ This is wild! ${}$ $\endgroup$
    – Unit
    Jul 31 '19 at 20:07
  • $\begingroup$ How would you adapt this method if $n^\varepsilon$ were replaced by $\log n$? $\endgroup$
    – Unit
    Aug 1 '19 at 18:07
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    $\begingroup$ Using $\ln n=\displaystyle\int_0^1\frac{1-t^{n-1}}{-\ln t}\,dt$ (essentially Frullani integral), one arrives at $$\sum_{n=2}^{\infty}\frac{x^n}{n!}\ln n=\underbrace{\gamma+\int_0^\infty\frac{e^{-t}\,dt}{x+t}}_{\text{small when }x\to\infty}+e^x\left(\ln x-x\int_0^1 e^{-xt}\ln\Big(\frac{1}{t}\ln\frac{1}{1-t}\Big)\,dt\right).$$ Expansion of the double logarithm, and Watson's lemma, give $$\sum_{n=2}^{\infty}\frac{x^n}{n!}\ln n\asymp e^x\left(\ln x-\frac{1}{2x}-\frac{5}{12x^2}-\frac{3}{4x^3}-\frac{251}{120x^4}-\ldots\right)$$ $\endgroup$
    – metamorphy
    Aug 4 '19 at 19:45
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This is a sort of limit theorem for the Poisson distribution.

Note that $f(\lambda) =\sum_{n=1}^\infty n^\varepsilon \frac{\lambda^n}{n!} =\mathbb{E}[g(X)]e^{\lambda}$ for $g(x):=x^{\epsilon}$ and a Poisson random variable $X$ with mean $\lambda$.

Lemma: $$|\mathbb{E}[g(X)]-g(\mathbb{E}[X])|\leq |\epsilon(\epsilon-1)|/2\lambda^{\epsilon-1}$$ Proof: $|g(x)-g(\lambda)-g'(\lambda)(x-\lambda)|\leq |g''(\lambda)/2|(x-\lambda)^2=|\epsilon(\epsilon-1)|\lambda^{\epsilon-2}/2(x-\lambda)^2$. Hence $$ |\mathbb{E}[g(X)]-g(\mathbb{E}[x])|=|\mathbb{E}[g(X)-g(\lambda)-g'(\lambda)(X-\lambda)]|\leq |\epsilon(\epsilon-1)|\lambda^{\epsilon-2}/2\mathbb{V}[X]\leq |\epsilon(\epsilon-1)|\lambda^{\epsilon-1}/2 $$ where we used that the variance of the Poisson distribution is $\lambda$.

Conclusion: $$f(\lambda)/(\lambda^{\epsilon}e^{\lambda})\to 1$$

Note that this works for all $\epsilon\in\mathbb{R}$.

You can also get more asymptotic terms by using more terms in the Taylor expansion of $g$ and using known formulas for the centered moments of the Poisson distribution.

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  • $\begingroup$ Thanks for this! It seems readily generalizable to other functions $g$. How did you get the first equality of absolute values after the word "Hence"? $\endgroup$
    – Unit
    Jul 31 '19 at 21:02
  • $\begingroup$ very appreciable intuition that of using the Poisson distribution (+1) $\endgroup$
    – G Cab
    Jul 31 '19 at 21:16
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    $\begingroup$ Ah! $\mathbb{E}[X-\lambda] = 0$. $\endgroup$
    – Unit
    Jul 31 '19 at 21:18
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    $\begingroup$ Actually, I don't see how the Taylor remainder is $|g''(\lambda)/2|(x-\lambda)^2$. I get that $|g''(x)| = \epsilon(1-\epsilon)x^{\epsilon-2}$ is decreasing, so if $x$ is to the right of $\lambda$ then we're set. But don't you need the inequality to hold for all $x$? $\endgroup$
    – Unit
    Jul 31 '19 at 21:44
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    $\begingroup$ @unit good point. The inequality is actually not true as stated. For $0<epsilon<1$ I believe it might be true for large enough $\lambda$ but I don't have a proof for that either. In general, you can prove that the inequality only needs to be true locally using a large deviation bound for the Poisson distribution (such as the Chernoff bound in the Wikipedia article), but that feels like to heavy a hammer. I'll see if I can find an easy fix. $\endgroup$
    – Bananach
    Aug 1 '19 at 5:13

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