Can we have any closed form or approximation of the summation? And can we know when it is positive?
$S(n,t)=\sum_{i=0}^{t}(-1)^{i}\tbinom{n}{i}$
We can assume that $0<t<n/2$.
This is the sum of the $D(n,t)=\tbinom{n}{t}-\tbinom{n}{t+1}$. A simpler form of $D(n,t)$ may help the problem.
Till now, I can not find any help from Stirling's approximation.
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It is well known that your sum $S(n,t) = (-1)^t \binom{n-1}{t}$. There are a variety of proofs here for instance. Among other things, this means that the sum is nonnegative if and only if $t$ is even.
As for asymptotics, there are lots of known approximations for $\binom{n-1}{t}$ which depend on how quickly $t$ and $n$ are growing. One useful set of bounds is
$$ \frac{n^k}{k^k} \leq \binom{n}{k} \leq \frac{n^k}{k!} $$
Which one can easily use to approximate $S(n,t) = (-1)^t \binom{n-1}{t}$.
Asymptotically, one can use stirling's approximation to obtain the sharpest result which is always true as follows:
First expand the definition of the closed form of $S(n,t)$:
$$S(n,t) = (-1)^t \binom{n-1}{t} = (-1)^t \frac{(n-1)!}{t!(n-1-t)!}$$
Then apply the stirling approximation:
$$ (-1)^t \frac{(n-1)!}{t!(n-1-t)!} \sim (-1)^t \frac{\sqrt{2 \pi (n-1)}(n-1)^{n-1}/e^{n-1}}{\left ( \sqrt{2 \pi t} t^t / e^t \right ) \left ( \sqrt{2 \pi (n-1-t)} (n-1-t)^{n-1-t}/e^{n-1-t}\right )} $$
Now you can cancel and simplify this to
$$ (-1)^t \sqrt{\frac{n-1}{2 \pi t (n-1-t)}} \frac{(n-1)^{n-1}}{t^t (n-1-t)^{n-1-t}} $$
This expression, while correct, is slightly unwieldy. If you know more information about how $t$ grows (for instance, is $t \approx \sqrt{n}$? $kn$? $n^{0.1}$?) you can simplify the above expression (sometimes dramatically). Google has lots of results in this vein.
I hope this helps ^_^
