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How could we prove that $$\lim_{x\to1^-}~\bigg(\sqrt[a]{1-x}\cdot\sum_{n=0}^\infty~x^{n^a}\bigg)~=~\Gamma\bigg(1+\frac1a\bigg)$$ for $a>0$ ?


The inspiration came to me while trying find a solution to this related question, which treats the

special case $a=2$. I am pretty confident that the best way to approach this is by some clever

manipulation of the well-known identities $~\displaystyle\int_0^\infty e^{-t^a}~dt~=~\Gamma\bigg(1+\frac1a\bigg)~$ and $~e^u~=~\displaystyle\sum_{n=0}^\infty\frac{u^n}{n!}~,$

but unfortunately I haven't been able thus far to capitalize on my own idea. I also tried writing

x as $1-\epsilon$, and then expanding $(1-\epsilon)^{n^a}$ into its binomial series, but this endeavor has also been

proven worthless, or, at the very least, it appears to be so in my hands. Which is where you, dear

reader, come in! Can you help me out of my impasse ? Any ideas or suggestions are welcome!

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1 Answer 1

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Use this well known : Let $f(x)$ monotonic function on $x\ge 0$,and this integeral $\int_{0}^{+\infty}f(x)dx$ exsit.then we have $$\lim_{h\to 0^{+}}h[f(h)+f(2h)+\cdots]=\int_{0}^{\infty}f(x)dx$$

then let $x=e^{-h^a}$,then \begin{align*}\lim_{x\to 1^{-}}\sqrt[a]{1-x}(1+x^{1^a}+x^{2^a}+\cdots+x^{n^a}+\cdots)&= \lim_{h\to 0^{+}}\dfrac{\sqrt[a]{1-e^{-h^a}}}{h}\cdot h(1+e^{-h^a}+e^{-(2h)^a}+\cdots)\\ &=\int_{0}^{+\infty}e^{-t^a}dt \end{align*}

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    $\begingroup$ Nice use of Riemann sum. +1 $\endgroup$ Mar 19, 2015 at 10:06
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    $\begingroup$ Brilliant ! Simply brilliant ! $\endgroup$
    – Lucian
    Mar 19, 2015 at 10:46
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    $\begingroup$ You are very welcome. $\endgroup$
    – math110
    Mar 19, 2015 at 10:47
  • $\begingroup$ @math110 how on earth does someone see this solution? $\endgroup$ Jan 23, 2017 at 21:41
  • $\begingroup$ @Euler_Salter Riemann sums... $\endgroup$ Jun 20, 2017 at 19:54

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