how to proof this Stirling related equation here is what I need to proof, have no idea were to start. I know there is some connection with the Stirling theorem.
$$
\sum_{i=0}^{d}\binom{m}{i} \leq \left ( \frac{em}{d} \right )^{d}
$$
I tried to open the sum and start from the right side of the equation but nothing achieved.
Thanks in advance.
 A: First, observe that
$${m!\over(m-i)!} \le m^i $$
for $0\le i\le m$.  (The two sides are equal for $i=0$ and $1$.  Otherwise the inequality is strict:  The left hand side is a product of $i$ positive integers, none greater than $m$.)  It follows that, for $0\le i \le d \le m$, we have
$$\left({d\over m}\right)^d{m\choose i}= {d^i\over i!}\left({d\over m}\right)^{d-i}{m!\over m^i(m-i)!} \le {d^i\over i!}, $$
hence
$$\left({d\over m}\right)^d\sum_{i=0}^d {m\choose i} \le \sum_{i=0}^d {d^i\over i!} \lt \sum_{i=0}^\infty {d^i\over i!} =e^d. $$
The desired inequality follows.
(Note:  The OP did not explicitly assume that $d\le m$, but it's reasonable to assume he or she meant to.  In particular, if $d\gt em$, the OP's inequality is false.)
A: Assume that $d\le m$. Certainly we have $\binom{m}{d}\le (\frac{em}{d})^d$. From this one can deduce
$$
f(m,d)=\sum_{i=0}^d\binom{m}{i} \le \frac{1-r^{d+1}}{1-r}\binom{m}{d}\le \frac{1-r^{d+1}}{1-r}\left(\frac{em}{d}\right)^d
$$
for $r=\frac{d}{m-d+1},$  see Sum of 'the first k' binomial coefficients for fixed n . However, this is not good enough. One could use then a better upper bound for $\binom{m}{d}$ in the last inequality.
EDIT: I just saw that Gerhard Pasemann has a better solution with case distinction
$3d< m$ and $3d \ge m$.
A: Here is a simple approach.  The left hand side  (for fixed $m$) is always at most $2^m$, so when is the
right hand side bigger than $2^m$?  Rewriting $k = \frac{m}{d},$  this is the same as asking for which $k$ is
$ek > 2^k$?  By inspection or calculus, one has it true for $1 \leq k \leq 3$, so when $d$ is between
$m/3$ and $m$, the inequality holds.  For $d \lt m/3$, the right hand side (by Stirling) is larger than
$\sqrt{2\pi d}\binom{m}{d}$, which in turn is larger than twice the largest summand on the left hand side.
As has been noted elsewhere, this is an upper bound for the sum when $3d \leq m, $ showing the inequality holds for small $d$.
Gerhard "Ask Me About Rough Estimates" Paseman, 2013.05.17 
