Limit of sum of sequences at infinity

Question

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 $$L=\lim_{n\to \infty} (10 \sqrt [n]{a_n} +5\sqrt [n]{b_n})$$

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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.

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Edit:

By some other method I dont mean that just converting the Gamma function to factorial representation and then using Stirling's approximation

Answer 1

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 $k=3$.

Answer 2

There is a very simple method of evaluating $L$. Fort any bounded measurable function $f$ on $[0,1]$ we have $\lim (\int_0^{1} |f|^{p})^{1/p} \to \|f\|_{\infty}$ as $ p \to \infty$. Hence $a_n^{1/n} \to \max \{(1-x^{2}): 0\leq x \leq 1\}=1$. Similarly $b_n^{1/n} \to 1$. So $L=10+5=15$. Actually, th result I have quoted is easy to prove in this case. Note that $\int_0^{1} (1-x^{2})^{n} \leq 1$. Also $\int_0^{1} (1-x^{2})^{n} \geq \int_0^{t} (1-x^{2})^{n} \geq t(1-t^{2})^{n}$ for each $t$. Take n-th root and conclude that $\lim \inf a_n^{1/n} \geq 1-t^{2}$ for anty $t>0$. Let $t \to 0$.