Given two sequences $$a_n=\int_0^1 (1-x^2)^n dx$$ and $$b_n=\int_0^1 (1-x^3)^n dx$$ ,($n\in N$) then find the value of $$L=\lim_{n\to \infty} (10 \sqrt [n]{a_n} +5\sqrt [n]{b_n})$$
My try:
$$a_n=\int_0^1 (1-x^2)^n dx= \frac 12 B\left( n+1, 1/2\right)=\frac 12\cdot \frac {\Gamma(n+1)\Gamma(1/2)}{\Gamma\left(n +\frac 32\right)}$$
Similarly $$b_n=\frac 13\cdot \frac {\Gamma(n+1)\Gamma(1/3)}{\Gamma\left(n +\frac 43\right)}$$
Now let's consider $$J=\lim_{n\to \infty} (b_n)^{1/n} =\lim_{n\to \infty} \left(\frac 13\right)^{1/n}\cdot \left(\frac {\Gamma(n+1)\Gamma(1/3)}{\Gamma\left(n +\frac 43\right)}\right) ^{1/n}$$ $$=\lim_{n\to\infty} \left(\frac 13\right)^{1/n}\cdot \left(\lim_{n\to \infty} \left(\frac {\Gamma(n+1)\Gamma(1/3)}{\Gamma\left(n +\frac 43\right)}\right) ^{1/n}\right)$$ $$=1\cdot \lim_{n\to \infty} \left(\frac {\Gamma(n+1)\Gamma(1/3)}{\Gamma\left(n +\frac 43\right)}\right) ^{1/n}$$ $$=\exp \left( \lim_{n\to\infty} \frac {\ln(\Gamma(n+1)) +\ln(\Gamma(1/3))-\ln(\Gamma(n +4/3))}{n}\right)$$
Now by L'Hospital we have $$\exp {\left( \lim_{n\to \infty} \frac {\ln(\Gamma(n+1)) +\ln(\Gamma(1/3))-\ln(\Gamma(n +4/3))}{n}\right)} =\exp{\left( \lim_{n\to \infty} \left[\psi(n+1) -\psi(n+ 4/3)\right] \right)}$$
But for large $x$, $\psi(x)\sim \ln (x)$ Hence $$\exp {\left( \lim_{n\to \infty} \left[\psi(n+1) -\psi(n+ 4/3)\right]\right)}=\exp {\left( \lim_{n\to \infty} \left[\ln(n+1) -\ln(n+ 4/3)\right]\right)}=1$$
Similarly $$\lim_{n\to \infty} (a_n)^{1/n}= 1$$
Hence giving me the answer as $15$.
I want to know whether I am correct or not. And if I am correct I want to know if there is any other simple and elementary method to solve the question.
Edit:
By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation
Best Answer
A much simpler approach. Let $k$ be a positive integer. Then for $x\in [0,1]$, $$1-x\leq 1-x^k\leq 1$$ and therefore $$\frac{1}{n+1}=\int_0^1 (1-x)^n dx\leq a_n(k):=\int_0^1 (1-x^k)^n dx\leq 1.$$ Hence, as $n\to\infty$, $$1\leftarrow\frac{1}{\sqrt[n]{n+1}}\le\sqrt[n]{a_n(k)}\leq \sqrt[n]{1}=1$$ and by the Squeeze Theorem it follows that $\sqrt[n]{a_n(k)}\rightarrow 1$. In your case, we need the limit for $k=2$ and $k=3$.