Limit of a sequence defined as a definite integral Given two sequences $$a_n=\int_{0}^{1}(1-x^2)^ndx$$ and $$b_n=\int_{0}^{1}(1-x^3)^ndx$$ where $n$ is a natural number, then find the value of $$\lim_{n \to \infty}(10\sqrt[n]{a_n}+5\sqrt[n]b_n)$$
I have no starts. Looks good though. Some hints please. Thanks.
 A: First $a_n\le1$. Fix $\delta\in(0,1)$. In $[0,\delta]$, $(1-x^2)^n$ is decreasing and hence
$$ \int_{0}^{1}(1-x^2)^ndx\ge\int_{0}^{\delta}(1-x^2)^ndx\ge(1-\delta^2)^n\delta. $$
So 
$$ (1-\delta^2)^n\delta\le a_n\le 1 $$
from which one has
$$ (1-\delta^2)\sqrt[n]\delta\le \sqrt[n]{a_n}\le 1. $$
Thus
$$ \lim_{n\to\infty}\sqrt[n]{a_n}=1.$$
You can use the same way to obtain 
$$ \lim_{n\to\infty}\sqrt[n]{b_n}=1. $$
So
$$\lim_{n \to \infty}(10\sqrt[n]{a_n}+5\sqrt[n]b_n)=15. $$
A: It is interesting to note that we can calculate the integrals explicitly. We have $$\begin{align} a_{n}=
  & \int_{0}^{1}\left(1-x^{2}\right)^{n}dx
 \\ =
  &\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\int_{0}^{1}x^{2k}dx
  \\ =
  & \sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{2k+1}\tag{1}
 \end{align}$$ and in a similar manner we have $$b_{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{3k+1}\tag{2}
 $$ and now we recall the 

Melzak's formula: Let $f(x)$ be a polynomial in $x$ of degree $n$, let $y$ be an arbitrary complex number. Melzak's formula states that for $y \ne 0,-1,-2,\ldots,-n$ \begin{align*} f(x+y)&=y\binom{y+n}{n}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{f(x-k)}{y+k}\\ \end{align*}

(for a proof see here) so for the calculus of $(1)$ we have, taking $f\left(x\right)=1$ and $y=1/2$ we get $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{2k+1}=\frac{1}{\dbinom{\frac{1}{2}+n}{n}}
 $$ and taking $f\left(x\right)=1$ and $y=1/3$ $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{3k+1}=\frac{1}{\dbinom{\frac{1}{3}+n}{n}}
 $$ and now using the Stirling's approximation we get $$\sqrt[n]{a_{n}}\rightarrow1,\,\sqrt[n]{b_{n}}\rightarrow1
 $$ and so 

$$\lim_{n\rightarrow\infty}\left(10\sqrt[n]{a_{n}}+5\sqrt[n]{b_{n}}\right)=\color{red}{15}.$$

A: We may notice that the power series
$$ \sum_{n\geq 0} a_n \xi^n  = \int_{0}^{1}\frac{dx}{1-\xi(1-x^2)}= \frac{1}{\sqrt{\xi(1-\xi)}}\,\arctan\left(\sqrt{\frac{\xi}{1-\xi}}\right)$$
has a radius of convergence equal to one, as well as the power series $\sum_{n\geq 0}b_n \xi^n$.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\mbox{Asymptotically}:
\\[3mm] &\ \braces{%
\left.\int_{0}^{1}\pars{1 - x^{\mu}}^{n}\,\dd x\,
\right\vert_{\ \mu\ >\ 0}}^{1/n} \sim {\alpha^{1/n}\pars{\mu} \over n^{1/\pars{\mu n}}}\
\stackrel{n\ \to \infty}{\longrightarrow}\
\color{#f00}{\large 1}\,,\qquad \mu = 2,3.
\\[3mm] &\ \mbox{where}\ \alpha\pars{\mu} >0\,, \forall\ \mu > 0
\\[3mm] & \mbox{}
\end{align}
In another words, $\ds{\lim_{n \to \infty}a_{n}^{1/n} = \lim_{n \to \infty}b_{n}^{1/n} = \color{#f00}{\large 1}}$.

$$
\mbox{Then,}\quad
\lim_{n \to \infty}\pars{\color{#f00}{10}\,a_{n}^{1/n} + \color{#f00}{5}\,b_{n}^{1/n}} = 10 + 5 = \color{#f00}{15}
$$
A: In the same spirit as previous answers, using the gamma function, we should get $$a_n=\frac{\sqrt{\pi }}2\frac{ \Gamma (n+1)}{ \Gamma \left(n+\frac{3}{2}\right)}$$ $$b_n=\Gamma \left(\frac{4}{3}\right)\frac{ \Gamma (n+1)}{\Gamma
   \left(n+\frac{4}{3}\right)}$$ Taking logarithms and using Stirling approximation $$\log(a_n)=\left(\frac{1}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi
   }}{2}\right)\right)+O\left(\frac{1}{n^{2}}\right)$$ which makes $$\sqrt[n]{a_n}=1+\frac{\frac{1}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi
   }}{2}\right)}{n}+O\left(\frac{1}{n^{2}}\right)$$ Similarly $$\log(b_n)=\left(\frac{1}{3} \log \left(\frac{1}{n}\right)+\log \left(\Gamma
   \left(\frac{4}{3}\right)\right)\right)+O\left(\frac{1}{n^{2}}\right)$$ which makes $$\sqrt[n]{b_n}=1+\frac{\frac{1}{3} \log \left(\frac{1}{n}\right)+\log \left(\Gamma
   \left(\frac{4}{3}\right)\right)}{n}+O\left(\frac{1}{n^{2}}\right)$$ So, $$\alpha \sqrt[n]{a_n}+\beta \sqrt[n]{b_n}=(\alpha +\beta )+\frac{ \alpha  \log \left(\frac{\sqrt \pi }{2}\right)+\beta  \log
   \left(\Gamma \left(\frac{4}{3}\right)\right)}{n}-\frac{(3 \alpha +2 \beta ) \log (n)}{6 n}+O\left(\frac{1}{n^{2}}\right)$$ which shows the limit and how it is approached.
