I'd like to compute $\sum_{i=0}^k {{N}\choose{i}}$. Is there a computable approximation for that?

2$\begingroup$ At first change your sigma to a definite integral by method of this post and after that use some approximation method about the definite integral. $\endgroup$ – Amin235 Feb 5 '17 at 12:31

7$\begingroup$ Possible duplicate of Sum of 'the first k' binomial coefficients for fixed n $\endgroup$ – Dirk Feb 5 '17 at 15:39
One of the more convenient and popular approximations of the sum is
$$\frac{2^{nH(\frac{k}{n})}}{\sqrt{8k(1\frac{k}{n})}} \leq \sum_{i=0}^k\binom{n}{i} \leq 2^{nH(\frac{k}{n})}$$
for $0< k < \frac{n}{2}$, where $H$ is the binary entropy function. (The upper bound is exactly what Aryeh Kontorovich mentions.) You can find its proof in many textbooks, but probably I first learned it from Chapter 10 of The Theory of ErrorCorrecting Codes by MacWilliams and Sloane.
Also, this post on MO asks a similar question and is a good resource for the sum in my opinion. You can find several other useful bounds there.

$\begingroup$ Nice, I didn't know about the upper bound. That's probably as well as you can do, with such simple estimates. $\endgroup$ – Aryeh Kontorovich Feb 5 '17 at 15:41

A wellknown upper bound, for $k\le N/2$, is $$ \sum_{i=0}^k {N\choose i} \le 2^{N H(k/N)},$$ where $H$ is the binary entropy function $$ H(x) = x\log_2(x)(1x)\log_2(1x).$$ This bound was sharpened in Lemma 5 of http://www.sciencedirect.com/science/article/pii/S0012365X12000544

$\begingroup$ The sharpening adds a prefactor of $0.98$; the authors comment that a smaller prefactor is possible with a more careful/detailed proof. The paper is also available, open access, on arXiv: arxiv.org/abs/1007.4915; there it is Lemma 7.1, in Section 7 $\endgroup$ – Sam T Aug 6 '19 at 9:59