Why is $\int_{0}^{1}{(1+x^2)^n dx} \sim \frac{2^n}{n} $? By $a_n \sim b_n$ I mean that $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 1$.
I don't know how to do this problem. I have tried to apply binomial theorem and I got
$$\int_{0}^{1}{(1+x^2)^n dx} = \int_0^1 \sum_{k=0}^n{\binom{n}{k}x^{2k}dx} = \sum_{k=0}^n \int_0^1{ \binom{n}{k}x^{2k}dx} = \sum_{k=0}^n  \frac {\binom{n}{k}}{2k+1}$$
But I don't know what I could do with this, nor if it is a correct approach. 
 A: Let $u=(1+x^2)/2$, so that $du=x\,dx=\sqrt{2u-1}\,dx$. It follows that
$${n\over2^n}\int_0^1(1+x^2)^n\,dx=n\int_{1/2}^1{u^n\over\sqrt{2u-1}}\,du=n\int_{1/2}^1u^n\,du+n\int_{1/2}^1u^n\left({1\over\sqrt{2u-1}}-1\right)\,du$$
Now
$$n\int_{1/2}^1u^n\,du={n\over n+1}\left(1-\left(1\over2\right)^{n+1}\right)\to1(1-0)=1$$
So it remains to show that
$$n\int_{1/2}^1u^n\left({1\over\sqrt{2u-1}}-1\right)\,du\to0$$
Note that
$$0\le{1\over\sqrt{2u-1}}-1={1-\sqrt{2u-1}\over\sqrt{2u-1}}={2(1-u)\over\sqrt{2u-1}(1+\sqrt{2u-1})}\le{1-u\over\sqrt{2u-1}}$$
so it's enough to show that
$$n\int_{1/2}^1{u^n(1-u)\over\sqrt{2u-1}}\,du\to0$$
It's convenient to let $u=1-v$ and, taking $n=N^3$ to be sufficiently large, rewrite the integral as
$$\begin{align}
N^3\int_0^{1/2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}}
&=N^3\int_0^{1/N^2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}}
+N^3\int_{1/N^2}^{1/2}{v(1-v)^{N^3}\,dv\over\sqrt{1-2v}}\\
&\le{N^3\over\sqrt{1-1/N^2}}\int_0^{1/N^2}v\,dv+{N^3\over2}\left(1-{1\over N^2}\right)^{N^3}\int_0^{1/2}{dv\over\sqrt{1-2v}}\\
&={1\over2N\sqrt{1-2/N^2}}+{N^3\over2}\left(\left(1-{1\over N^2}\right)^{N^2} \right)^N\\
&\le{1\over N}+N^3(e^{-1/2})^N\\
&\to0+0=0
\end{align}$$
(Note, these calculations do not require $n$ to be an integer.)
A: A partial progress
We make the substitution $x\tan(y)$. Then $dx=\sec^2(y)dy$.
Now $$\int_{0}^{1}(1+x^2)^ndx=\int_{0}^{\frac{\pi}{4}}\sec^{2n}(y)\sec^2(y)dy=\int_{0}^{\frac{\pi}{4}}\sec^{2(n+1)}(y)dy$$
Maybe it will be helpful.
Here is the proof by proceeding in your way.
$$\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2k+1}>\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2k+2}=\frac{n}{2(n+1)}\sum_{k=0}^{n}\frac{n+1}{k+1}\binom{n}{k}=\frac{n}{2(n+1)}\sum_{k=0}^{n}\binom{n+1}{k+1}=\frac{n}{2(n+1)}(2^{n+1}-1)$$
Hence $$\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2^n(2k+1)}\geq\lim_{n\rightarrow\infty}\frac{n}{2(n+1)}\left(\frac{2^{n+1}}{2^n}-\frac{1}{2^n}\right)=1$$
Again we see that,
$$\sum_{k=0}^{n}\frac{\binom{n}{k}}{2k+1}<1+\left(\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}\right)$$
From this you can show that $$\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{n\binom{n}{k}}{2^n(2k+1)}\leq1$$
Hence the result will follow.
A: You can also write $ \int_0^1 (1+x^2)^ndx = \int_0^1 e^{n\log(1+x^2)}dx $ and use a generalization of Laplace's method to handle the boundary case.
