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!


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*}

| cite | improve this answer | |
  • 4
    $\begingroup$ Nice use of Riemann sum. +1 $\endgroup$ – Start wearing purple Mar 19 '15 at 10:06
  • 2
    $\begingroup$ Brilliant ! Simply brilliant ! $\endgroup$ – Lucian Mar 19 '15 at 10:46
  • 2
    $\begingroup$ You are very welcome. $\endgroup$ – math110 Mar 19 '15 at 10:47
  • $\begingroup$ @math110 how on earth does someone see this solution? $\endgroup$ – Euler_Salter Jan 23 '17 at 21:41
  • $\begingroup$ @Euler_Salter Riemann sums... $\endgroup$ – Simply Beautiful Art Jun 20 '17 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.