# A double inequality with binomials

I don't see what I should use here. What would you use? $$\frac{2^{2013}}{2013}\le\frac{\binom{2013}{0}}{1}+\frac{\binom{2013}{1}}{3}+\frac{\binom{2013}{2}}{5}+\cdots+\frac{\binom{2013}{2013}}{2\cdot 2013+1}\le\frac{2^{2013}}{2012}$$

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Are you trying to prove the inequality? –  Jonathan Rich Mar 19 '13 at 20:24
@JonathanRich: what else I could do with it? Perhaps to admire it. –  James the son of Zebedee Mar 19 '13 at 20:53
Perhaps it may be useful to let $f(x)=\sum \binom{2013}{k}x^{2k}=(1+x^2)^{2013}$. Integrate term by term from $x=0$ to $x=1$. –  André Nicolas Mar 19 '13 at 21:08
@AndréNicolas : I had the same idea, but I don't how to bound $\int_0^1 \left(1+u^2\right)^n \, du$ correctly. Using $(2u)^n \le \left(1+u^2\right)^n \le (1+u)^n$ yields $$\frac{2^n}{n+1} \le \sum_{k = 0}^n \frac{\binom{n}{k}}{2k+1} \le \frac{2^{n+1}-1}{n+1}$$ –  Joel Cohen Mar 19 '13 at 21:16
@JoelCohen: I scribbled quickly, using my favourite quickie estimate tool, integration by parts. Seems to work. Maybe. –  André Nicolas Mar 19 '13 at 21:31

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.

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+1. Glad to see you are back to answering. Welcome back! –  user17762 Mar 29 '13 at 2:15
@Marvis: Thanks! Hopefully I can increase my participation... –  Aryabhata Mar 29 '13 at 2:17