A comparison on binomial expansion

[forumfann]

Could anyone help me on this question? Is it true that
\sum_{k=n+1}^{2n}\left(\begin{array}{c}<br /> 2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}\leq2x
for any x\in(0,1) and any positive integer n?

Any help on that will be greatly appreciated!

 

[Gib Z]

Well, the LHS seems to be the "second half" of the terms of the expansion of ( x+ (1-x) )^(2n) = 1.

This gives that the LHS is 1 - \sum_{k=0}^n \left(\begin{array}{c}<br /> 2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}.

I can't see how that's incorrect, but then I note the inequality doesn't seem to hold anymore for when x is small, for the LHS would seem to tend to 1 whilst the RHS goes to zero. Maybe I'm going crazy..

 

[forumfann]

the LHS would seem to tend to 1

When k=0, x^k(1-x)^{2n-k} goes to 1 as x goes to 0, so 1-\sum_{k=0}^{n}\left(\begin{array}{c}<br /> 2n\\<br /> k\end{array}\right)x^{k}\left(1-x\right)^{2n-k} tends to 0 as x goes to 0 because the other terms with k>0 has a factor x. But thanks anyway.

 

[Gib Z]

Well if that is true, haven't you shown that the LHS is negative? I must have done something completely wrong as it seems my idea implies the original LHS must be negative..

Today obviously isn't my day. Sorry for wasting your time !

 

[forumfann]

The LHS is always positive for x\in(0,1). That is fine.

 

[forumfann]

In fact, I have checked the cases of n=1 through 1000, and it holds for all the cases. But I still don't know how to show it in general. Any other help?