TITLE: A double inequality with binomials
QUESTION [6 upvotes]: I don't see what I should use here. What would you use?
$$\frac{2^{2013}}{2013}\le\frac{\binom{2013}{0}}{1}+\frac{\binom{2013}{1}}{3}+\frac{\binom{2013}{2}}{5}+\cdots+\frac{\binom{2013}{2013}}{2\cdot 2013+1}\le\frac{2^{2013}}{2012}$$

REPLY [2 votes]: As noted by Andre, Integration by parts works.
If $$I_{n} = \int_{0}^{1} (1 + x^2)^n$$
by integration by parts, we get that
$$ I_{n} = x(1+x^2)^n \vert_{0}^{1} - \int_{0}^{1} x \frac{d(1+x^2)^n}{dx} $$
And some manipulations yields
$$I_{n} = \frac{2^n + 2(n+1)I_{n-1}}{2n+1}$$
And a straight-forward induction proof yields the upper bound 
$$I_{n} \le \frac{2^n}{n-1}$$
The lower bound was noted by Joel.