Find the limit of given expressions: Given sequences $$a_n=\int_0^1 (1-x^2)^n \,dx$$ and $$b_n=\int_0^1 (1-x^3)^n \,dx$$ where $n \in \mathbb{N}$, find $$\displaystyle\lim_{n\to \infty}(10\sqrt[n]{a_n}+5\sqrt[n]{b_n}).$$
So far:
I tried integrating expansions of $(1-x^2)^n$ and $(1-x^3)^n$ to get the series $C_0-\frac{C_1}{3}+\frac{C_2}{5}...\frac{(-1)^nC_n}{2n+1}$ and $C_0-\frac{C_1}{4}+\frac{C_2}{7}...\frac{(-1)^nC_n}{3n+1}$.
Since that was not working out, I tried integrating by parts:
$$ \begin{align} a_n=\int_{0}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{0}^{1}+2n\int_{0}^{1}x^2(1-x^2)^{n-1}dx \\\\&=0+2n\int_{0}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx \\\\&=2na_{n-1}-2na_{n} \end{align} $$ then, with $a_0=1,\,a_1=\frac23,$ $$ a_{n}=\frac{2n}{2n+1}\cdot a_{n-1}, \quad n\ge1. $$
Hence it comes out as $$a_n=\frac{1\cdot2\cdot4\cdot6...2n}{1\cdot3\cdot5\cdot7...2n+1}$$
Similarly integrating $b_n$, the result came out as:
$$b_n=\frac{3n}{3n+1}\cdot b_{n-1}, \quad n\ge1. $$
$$b_0=1,\,b_1=\frac34$$
Transforming into
$$b_n=\frac{1\cdot3\cdot6\cdot9...3n}{1\cdot4\cdot7\cdot10...3n+1}$$
Now I have to somehow put it into the given limit, but I have no idea how.
 A: There is no need to compute $a_n$ and $b_n$ explicitly to find the limit. Let $\epsilon >0$. Then $ 1\geq a_n \geq \int_0^{\epsilon} (1-x^{2})^{n} dx \geq (1-\epsilon^{2})^{n} \epsilon $. Taking $n-th$ roots , letting $n \to \infty$ and then $\epsilon \to 0$ we see that $a_n^{1/n} \to 1$ Simialrly, $b_n^{1/n} \to 1$. Hence the answer is $15$.
[ $1 \geq a_n^{1/n} \geq (1-\epsilon^{2}) \epsilon ^{1/n}$. Letting $ n \to \infty$ we get $1 \geq \lim \sup a_n^{1/n} \geq \lim \inf a_n^{1/n} \geq 1-\epsilon^{2}$. This is true for each $\epsilon>0$ so $a_n^{1/n} \to 1$].
A: you have worked really hard to tackle this problem. Now, I have some good news. This problem requires a lot less work. If you are familiar with the Dominated Convergence Theorem then I suggest you use it. Before stating it I would like to argue that your problem is one of proving that you can bring your limit within the integrals. Once you have done the letter the rest is history.
(1) Dominated COnvergence Theorem (DCT).
Let $(f_{n})$ be a sequence of complex-valued measurable functions on a measure space $(S,\Sigma, \mu)$. Suppose that the sequence converges pointwise to a function $f$ and is dominated by some integrable function $g$ in the sense that
$$|f_{n}(x)|\leq g(x)$$
for all numbers $n$ in the index set of the sequence and all points $x\in S$. Then $f$ is integrable (in the Lebesgue sense) and
$$\lim_{n\rightarrow \infty}\int_{S}|f_{n}-f|d\mu = 0 $$
which implies that
$$\lim_{n\rightarrow \infty}\int_{S}f_{n}d\mu = \int_{S}fd\mu$$

*

*For your problem you are working with the measure space $(S, \Sigma, \mu)$ where $S = [0,1]$ and $\Sigma$ and $\mu$ are the usual sigma alrgebra of Lebesgue measure one works with in regular integrals. The important thing to note is that the integrands $(1-x^{2})^{n}$ and $(1-x^{3})^{n}$ are both dominated by the constant integrable function $g(x) = 1$. i.e. $|(1-x^{2})^{n}|\leq 1$ and $|(1-x^{3})^{n}|\leq 1$ for all $n$. This means that

$$lim_{n\rightarrow \infty } a_{n} =\int_{0}^{1}\lim_{n\rightarrow \infty}a_{n}dx = \int_{0}^{1}1dx = 1$$
similarly
$$lim_{n\rightarrow \infty } b_{n} = 1.$$
Now that the hardest part is out of the way let us return to the problem at hand.
i.e. $\lim_{n\rightarrow \infty} \{10a_{n}^{\frac{1}{n}}+5b_{n}^{\frac{1}{n}}\}$
it takes little to convince yourself that
(2)$$10a_{n}+5b_{n}\leq10a_{n}^{\frac{1}{n}}+5b_{n}^{\frac{1}{n}}\leq 10+5 $$
the latter is simply due to the fact that the integrand can be no greater than one and the contracting property of the square root when applied to numbers in (0,1).
Taking the limit on al parts of the inequaltiy (2) we have
(3)$$\lim_{n\rightarrow\infty}\{10a_{n}+5b_{n}\}\leq\lim_{n\rightarrow\infty}\{10a_{n}^{\frac{1}{n}}+5b_{n}^{\frac{1}{n}}\}\leq 15$$
But using our results fromm earlier $\lim_{n\rightarrow\infty}\{10a_{n}+5b_{n}\}=10\lim_{n\rightarrow\infty}a_{n}+5\lim_{n\rightarrow\infty}b_{n}= 15$.
We therefore see that (3) is just
$$15\leq\lim_{n\rightarrow\infty}\{10a_{n}^{\frac{1}{n}}+5b_{n}^{\frac{1}{n}}\}\leq 15$$
which means that
$$\lim_{n\rightarrow\infty}\{10a_{n}^{\frac{1}{n}}+5b_{n}^{\frac{1}{n}}\} = 15.$$
This is not the most direct path home but it is a path nevertheless. I just notice a much shorter way to approach this given by @Kavi Rama Murthy and it looks easier on the eyes :).
