Can anyone prove (or disprove) the following?

Question

Let a(n) = C(2n,n) / (n+1)
where C(2n,n) = (2n)Cn = (2n)! / (n!)^2

Let f(x) = ∑ a(n) * x^(n+1) * (1-x)^n
for 0 < x < 1, where n goes from 0 to infinity

Can anyone prove (or disprove) that
1) f(x) = 1 (constant) when 0.5 < x < 1
2) f(x) = x / (1-x) for 0 < x < 0.5 ?

I'm sure you can tell that the second statement is true if the first one is true.

Answer 1

Let's first consider the interval 0 Notice that
f(x)=sum{n=0 to infinity}
(2n)! / (n!)^2/(n+1)*x^(n+1)*(1-x)^n
=1/(1-x)*sum{n=0 to infinity}
(2n)! /(n!)^2/(n+1)*x^(n+1)
*(1-x)^(n+1)
=1/(1-x)*sum{n=0 to infinity}
(2n)! / (n!)^2/(n+1)*(x*(1-x))^(n+1)
Now define the function
g(t)=sum{n=0 to infinity}
(2n)! / (n!)^2/(n+1)*t^(n+1)
then
f(x)=1/(1-x)*g(x*(1-x)).
Notice that
g'(t)=sum{n=0 to infinity}
(2n)! / (n!)^2*t^n
This series is well-known to converge to
1/sqrt(1-4*t) for 0<=t<1/4, so
g'(t)=1/sqrt(1-4*t).
Integrating with respect to t yields
g(t)=-1/2*sqrt(1-4*t)+C.
0=f(0)=1/(1-0)*g(0*(1-0))
=g(0), so
C=1/2:
g(t)=1/2-1/2*sqrt(1-4*t)
Now since
f(x)=1/(1-x)*g(x*(1-x)), we get
f(x)=1/(1-x)
*(1/2-1/2*sqrt(1-4*x*(1-x)))
=1/(1-x)*(1/2-1/2*(1-2*x))
=x/(1-x) for the interval 0<=x<=0.5.

For 0.5 f(x)/x=f(1-x)/(1-x). This implies
f(x)=x/(1-x)*f(1-x)
but for these x, 1-x lies in (0,0.5),
so we know that f(1-x)=(1-x)/(1-(1-x))
=(1-x)/x.
Substituting, we get
f(x)=x/(1-x)*(1-x)/x=1.


EDIT: Whoever gave me a thumbs down is invited to point out any error in my analysis.

Answer 2

Well a(n) (unless i'm much mistaken) is a catalan number...

which has a well known generating function...

(1-sqrt(1-4*x))/(2x) = sum(a(n)*x^n) for n = 0 to infinity

setting y = x(1-x) gives

f(x) = x * sum(a(n) y^n)
= (1-sqrt(1-4*x*(1-x))/(2*(1-x)) (after replacing y with x)

you can do the rest from this :-)

Answer 3

the first ataement is incorrect as
f(X) could be more than half and lower than 1 but not 0.5 or1
so the first one is incorrect also the second
one but if the first is true so the second is true
please don't forget giving me best answer

Answer 4

No I cannot.