Evaluating $\sum^n_{x=1}{2x-1\choose x}t^x$

Question

Is there any technique that I can use to evaluate $$\sum^n_{k=1}{2k-1\choose k}t^k, \quad \forall t\in\left(0,\frac{1}{4}\right)$$

It can be shown that the series converges even if $n\to \infty$ as $$ \frac{{2k+1\choose k+1}t^{k+1}}{{2k-1\choose k}t^k} = 2t \times \frac{2k+1}{k+1} < 4t < 1. $$ However, I can't find out how to simplify the summation.

Answer 1

We have:

$$\sum_{k=0}^n\binom{2k-1}kt^k=\frac12\sum_{k=0}^n\binom{2k}kt^k\underset{n\to\infty}\longrightarrow\frac1{\sqrt{1-4t}}$$

where the limit is well known as the generating function for the central binomial coefficient. One can rewrite the partial sums using hypergeometric functions, but I doubt anything simpler can be done.