I am interested in the following limit \begin{align} \lim_{t \to \infty} \sup_{x \ge 0} \frac{1}{x^2+1} e^{-x} \sum_{n\ge t} \frac{ n x^{n}}{n!}=0 \end{align}

Note that $\sum_{n>\ge} \frac{ n x^{n}}{n!} \le x e^{x}$. I also think that he following expression might be useful \begin{align} \sum_{n\ge t} \frac{ x^{n}}{n!} = \frac{\gamma(t,x)}{\Gamma(t)} \end{align} where $\gamma(t,x)$ is the incomplete gamma function. The expression for the incomplete gamma function canbe justifed from https://en.wikipedia.org/wiki/Incomplete_gamma_function where it is shown that \begin{align} \Gamma(t,x)=\Gamma(t) \sum_{n \le t-1} \frac{ x^{n}}{n!} \end{align} for ineteger $t$. Uisng the fact that $\gamma(t,x)=\Gamma(t)-\Gamma(t,x)$ concludes the proof.

  • $\begingroup$ Can you justify the expression of the sum? I didn't know it. $\endgroup$ Jul 15, 2018 at 19:27
  • $\begingroup$ @RafaelGonzalezLopez Sure. I just added the explanation and a reference. $\endgroup$
    – Lisa
    Jul 15, 2018 at 19:34

1 Answer 1


You don't need explicit expressions to prove the result. As you noticed,

$$f(x,t) := \frac{e^{-x}}{1+x^2} \sum_{n \geq t} \frac{x^n}{(n-1)!} \leq \frac{x}{1+x^2}.$$

For large values of $x$, the right hand-side is small. For small values, you get uniform convergence to $0$. To prove it, let $M \geq 0$. Then:

$$\sup_{x \geq 0} f(x,t) = \max \left\{ \sup_{x \in [0,M]} f(x,t), \sup_{x \geq M} f(x,t) \right\}.$$

Then, one the one hand:

$$\sup_{x \in [0,M]} f(x,t) \leq \sup_{x \in [0,M]} \frac{e^{-x}}{1+x^2} \cdot \sup_{x \in [0,M]} \sum_{n \geq t} \frac{x^n}{(n-1)!} \to_{t \to + \infty} 0,$$

and on the other hand:

$$\sup_{x \geq M} f(x,t) \leq \frac{M}{1+M^2}.$$

Using, for instance, a diagonal argument, you can prove what yo want.

  • $\begingroup$ Can you remind me what the diagonal argument is? $\endgroup$
    – Lisa
    Jul 15, 2018 at 20:28
  • $\begingroup$ In your bound, you have that the limit is bounded by $\max(0, \frac{M}{1+M^2}) $ for any $M$. So the next step to is to take $M$ to infinity or simply minimize over $M$ right? $\endgroup$
    – Lisa
    Jul 15, 2018 at 20:51
  • $\begingroup$ @Lisa: that's it. If you want a slightly cleaner argument, you get $0 \leq \liminf_t \leq \limsup_t \leq M/(1+M^2)$ for all $M$. That advantage is that you don't have to wonder about the existence of the limit. $\endgroup$
    – D. Thomine
    Jul 15, 2018 at 21:18

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