# Prove $$\lim_{n \to \infty} \int_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0$$

How to prove

\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0

was a question posed at Mathematics Stack Exchange. There are more efficient answers there, but here is a fun and interesting solution.

Using the integral defintion (analytically continued) of Gauss’s hypergeometric function

{}_{2}\mathrm{F}_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\limits_{0}^{1} t^{a-1} (1-t)^{c-b-1} (1-zt)^{-a} dt

for $$\mathrm{Re}\,c \gt \mathrm{Re}\,b \gt 0,\,\,|\mathrm{arg}(1-z)| \lt \pi$$

We have, using the substitution $$y=x^{n}$$
\begin{align}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx
&= \frac{1}{n} \int\limits_{0}^{1} \frac{y^{1/n}}{\sqrt{1+y}} dy \\
\tag{1}
&= \frac{\Gamma(1+1/n)\Gamma(1)}{n\Gamma(2+1/n)} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) \\
&= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
\end{align}

Taking the limit of the hypergeometric function and applying the integral definition again yields
\begin{align}
\lim_{n \to \infty} {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
&= {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1;2;-1 \right) \\
&= \frac{\Gamma(2)}{\Gamma(1)\Gamma(1)} \int\limits_{0}^{1} \frac{1}{\sqrt{1+t}} dt \\
&= 2(\sqrt{2} – 1)
\end{align}

Substituting this result into equation 1, we have

\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0*2(\sqrt{2} – 1) = 0