An asymptotic question Give $x \in (0,0.5)$, how can we compute the asymptotic result of $\sum_{k=n}^{\infty} {k+n \choose n} x^{k}$ as $n \rightarrow \infty$? Thanks. 
 A: To expand my previous comment: Recalling the binomial series $$(1-x)^{-n-1}=\sum_{k=0}^\infty\bigg({k+n\atop n} \bigg)x^k $$
your expression is the remainder relative to the Taylor polynomial of order $n-1$ centered at $0$ of the function $(1-x)^{-n-1}$. Thus by the integral formula of the remainder
$$R_n:=\sum_{k=n}^\infty\bigg({k+n\atop n} \bigg)x^k=n\bigg({2n\atop n} \bigg)\int_0^x(1-t)^{-2n-1}(x-t)^{n-1} dt \, .$$
To get an asymptotics for the integral, change variable linearly putting $nt=xs$, thus obtaining
$$R _ n=\bigg({2n\atop n} \bigg) x^ n  \int _ 0 ^ \infty \big(1-\frac{xs}{n}\big)^{-2n-1}\big(1-\frac{s}{n}\big) ^ {n-1} \chi_{[0,n]}(s)ds = $$
$$=  \bigg({2n\atop n} \bigg) x ^ n \int _ 0 ^ \infty e ^ {(2x-1)s}ds\big(1+o(1)\big)= \bigg({2n\atop n} \bigg) x ^ n \frac{ 1 }{1-2x}\big(1+o(1)\big) \, , $$
by the Lebesgue dominated convergence theorem (indeed, for large $n$ the integrands are dominated by $e^{cs}$ for any  $2x - 1  < c < 0 \, . $ 
A: The OP doesn't need this case so I'll just give a sketch. The values 
$$p_k = (1-x)^{n-1} x^k\binom{n+k}{n}$$
sum to 1 over $k\ge 0$, so they are the probabilities of a nonnegative integer random variable $X$.  Moreover, the probability generating function of $X~$ is
$$\sum_{k=0}^\infty p_k z^k = \left(\frac{1-x}{1-zx}\right)^{n+1},$$
so $X~$ is the sum of $n+1$ iid geometric random variables with probability generating function $(1-x)/(1-zn)$.  This means that $X~$ is asymptotically normal with
$$ \mathrm{E}(X) = \frac{x}{1-x}(n+1), \quad \mathrm{Var}(X) = \frac{x}{(1-x)^2}(n+1).$$
The truncated sum requested by the OP is $(1-x)^{1-n}\mathrm{Prob}(X\ge n)$.  The normal approximation will give a good answer if $x$ is close enough to $\frac12$ (details left for the reader).
A: Let $f(x,n) = \sum_{k=n}^\infty {k+n \choose n} x^k$.
For given $x$ and $n$, the largest value of $g(k) = {k+n \choose n} x^k$ is at $k = \frac{x}{1-x} n - 1$, and $g(k)$ is increasing below this and decreasing above.  If $x \le 1/2$, $f(x,n)$ is dominated by the terms near $k=n$, and I think the result will be
$f(x,n) \sim {2n \choose n} \frac{x^n}{1-2x}$ if $x < 1/2$.  If $x > 1/2$, 
$f(x,n)$ is dominated by terms with $k > n$, and I think the result will be
$f(x,n) \sim \frac{1}{(1-x)^{n+1}}$
A: Concerning the case $x\rightarrow \frac{1}{2}$ we have :
$f(1/2,n)= 2^n+\binom{2n}{n} 2^{-n}$   (nearly $2^n (1+1/\sqrt{\pi n}))$
found at page 247 of Concrete Mathematics Graham, Knuth, and Patashnik. See too equation (5.20) and the discussion in 'partial sum involving factorials' of stackexchange.
(this should have been a comment anyway...)
