Binomial fraction sum to infinity Compute the limit:
$$\lim_{n\to\infty} \sum_{k=0}^n \frac {\dbinom{n}{k}}{\dbinom{2n-1}{k}}$$
Here i tried to give some k values to the sum hoping to see a possible pattern, 
but i didn't figure out any such a pattern. I wonder if there is an easy way to
solve such a limit.
 A: First expand the fraction:
$$\begin{align*}
\frac {\dbinom{n}{k}}{\dbinom{2n-1}{k}}&=\frac{n!k!(2n-1-k)!}{k!(n-k)!(2n-1)!}\\
&=\frac{n!(2n-1-k)!}{(n-k)!(2n-1)!}\\
&=\frac{n!}{(2n-1)!}\cdot\frac{(2n-1-k)!}{(n-k)!}\;.
\end{align*}$$
Thus,
$$\begin{align*}
\lim_{n\to\infty} \sum_{k=0}^n \frac {\dbinom{n}{k}}{\dbinom{2n-1}{k}}&=\lim_{n\to\infty}\frac{n!}{(2n-1)!}\sum_{k=0}^n\frac{(2n-1-k)!}{(n-k)!}\\\\
&=\lim_{n\to\infty}\frac{n!}{(2n-1)!}\sum_{k=0}^n\frac{(n-1+k)!}{k!}\\\\
&=\lim_{n\to\infty}\frac{n!(n-1)!}{(2n-1)!}\sum_{k=0}^n\frac{(n-1+k)!}{k!(n-1)!}\\\\
&=\lim_{n\to\infty}\binom{2n-1}{n}^{-1}\sum_{k=0}^n\binom{n-1+k}{n-1}\\\\
&=\lim_{n\to\infty}\binom{2n-1}{n}^{-1}\binom{2n}n\\\\
&=\lim_{n\to\infty}\frac{n!(n-1)!(2n)!}{(2n-1)!n!^2}\\\\
&=\lim_{n\to\infty}\frac{2n}n\\\\
&=2\;.
\end{align*}$$
Note that the sum is $2$ for all $n>0$, so the limits in the problem statement and the calculation are superfluous, though I discovered this only at the end of the calculation, when it came as a pleasant surprise. If I wanted to present a polished version of the argument, I would go back and remove ‘$\lim\limits_{n\to\infty}$’ everywhere that it appears.
Added: This result has a rather nice interpretation. The $k=0$ term in the summation is always $1$, so we’ve shown that $$\sum_{k=1}^n\frac{\dbinom{n}k}{\dbinom{2n-1}k}=1$$ for $n>0$. Now imagine that you have a box of $2n-1$ marbles, identical save for their color: $n$ are black, and $n-1$ are white. You perform $n$ independent experiments $E_1,\dots,E_n$ determining the values of $n$ random variables $X_1,\dots,X_n$ respectively. $E_k$ consists in drawing a random sample of $k$ marbles from the box; the experiment is a success if the sample consists entirely of black marbles, in which case $X_k=1$; otherwise, if the sample contains at least one white marble, $X_k=0$. Clearly $$\Bbb E(X_k)=\mathrm{Pr}(X_k=1)=\frac{\dbinom{n}k}{\dbinom{2n-1}k}\;.$$ Now let $X=\sum\limits_{k=1}^nX_k$, the number of successes in the $n$ experiments taken together. Then
$$\Bbb E(X)=\sum_{k=1}^n\Bbb E(X_k)=\sum_{k=1}^n\frac{\dbinom{n}k}{\dbinom{2n-1}k}=1\;,$$
meaning that on average we can expect one success among the $n$ experiments.
(This would be even nicer if I could see a good intuitive reason to expect one success on average!)
