An upper bound for $\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$? Does anyone know of a reasonable upper bound for the following:
$$\sum_{i = 1}^m \frac{\binom{i}{k}}{2^i},$$
where we $k$ and $m$ are fixed positive integers, and we assume that $\binom{i}{k} = 0$ whenever $k > i$.
One trivial upper bound uses the identity $\binom{i}{k} \le \binom{i}{\frac i2}$, and the fact that $\binom{i}{\frac{i}{2}} \le \frac{2^{i+1}}{\sqrt{i}}$, to give a bound of 
$$2\sum_{i = 1}^m \frac{1}{\sqrt{i}},$$
where $\sum_{i = 1}^m \frac{1}{\sqrt{i}}$ is upper bounded by $2\sqrt{m}$, resulting in a bound of $4\sqrt{m}$. Can we do better?
Thanks!
Yair

I like Pot's argument better, but here's another approach for those who may be interested.

I get an upper bound of $2$, independent of $k$ and $m$.  
Applying summation by parts, we have, for $k \geq 2$,
$$\begin{align*}
\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} &= \binom{m}{k} \sum_{i=1}^m \frac{1}{2^i} - \sum_{i=1}^{m-1} \left(\binom{i+1}{k} - \binom{i}{k}\right) \sum_{j=1}^i \frac{1}{2^j} \\
&= \binom{m}{k} \left(1 - \frac{1}{2^m}\right) - \sum_{i=1}^{m-1} \binom{i}{k-1} \left(1 - \frac{1}{2^i}\right)\\
&= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \sum_{i=1}^{m-1} \binom{i}{k-1} + \sum_{i=1}^{n-1} \binom{i}{k-1} \frac{1}{2^i}\\
&= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \binom{m}{k} + \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}\\
&= \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}- \binom{m}{k}\frac{1}{2^m},
\end{align*}$$
where, in the second-to-last step, we use the upper summation identity for the binomial coefficients, $\displaystyle \sum_{i=0}^m \binom{i}{k} = \binom{m+1}{k+1}$. 
Letting $\displaystyle F(m,k) = \sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$, this means we have the recurrence $$F(m,k) = F(m-1,k-1) - \binom{m}{k}\frac{1}{2^m},$$ valid for $k \geq 2$.
Unrolling the recurrence is easy, and with $F(m-k+1,1) = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}},$ we have
$$\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}} - \sum_{i=0}^{k-2} \binom{m-i}{k-i} \frac{1}{2^{m-i}}.$$
Therefore, 
$$\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} \leq 2.$$
