Lower Bound for Bernstein Approximation If have to do solve the following problem:
Let $f(t) = | t - \frac{1}{2} |$ be defined on $[0,1]$ and let $B_n ( t )$ denote the $n$-th Bernsteinpolynomial for $f$, i.e.
$$
B_n ( t ) =
\sum_{i = 0}^n \binom{n}{i} t^i (1 - t)^{n - i} f\bigg(\frac{i}{n}\bigg)
\text{.}
$$
Using Stirling's formula I have to show, that there exists a $c > 0$ which is independent from $n$, such that
$$
\bigg| f \bigg( \frac{1}{2} \bigg) - B_n \bigg( \frac{1}{2} \bigg) \bigg| >
\frac{c}{\sqrt n}
\text{.}
$$
Our professor also pointed out, that we must not use Stirling's formula to approximate $i!$ in the binomial coefficient contained in $B_n$, since this approximation could be far off as $i$ runs through low values.
Since
$$
\bigg| f \bigg( \frac{1}{2} \bigg) - B_n \bigg( \frac{1}{2} \bigg) \bigg| =
\bigg| B_n \bigg( \frac{1}{2} \bigg) \bigg|
$$
I suppose, that there is a clever way to simplify $B_n$, but I couldn't find a useful way to do this so far. Could anyone give me a hint in order to point me to the right direction?
I appreciate any help!!

$
\newcommand{\floor}[1]{\left\lfloor{#1}\right\rfloor}
$
Hint: A bit of simplification is needed before we can apply Stirling's formula. Notice that
\begin{align}B_n(1/2) &= \frac{1}{2^n} \sum_{i=0}^n \binom{n}{i} \left|\frac{i}{n} - \frac12\right| = \frac{2}{2^{n}} \sum_{i=0}^{\floor{n/2}} \binom{n}{i}\left(\frac12 - \frac{i}{n}\right)\\
&=\frac{1}{2^n} \sum_{i=0}^{\floor{n/2}}\binom{n}{i} - \frac{1}{n2^{n-1}}\sum_{i=0}^{\floor{n/2}} \binom{n}{i}i
\end{align}
In order to tackle these sums, I suggest you handle the cases $n = 2k$ and $n = 2k+1$ separately; but armed with the knowledge that $\sum_{i=0}^n \binom{n}{i} = 2^n$ and that $\sum_{i=0}^n \binom{n}{i}i = 2^{n-1}n$, you should be able to figure them out (extra hint: in the second sum, remember that $\binom{n}{n-i}(n-i) = n\binom{n}{i} - i \binom{n}{i}$ and use the value of the first sum).
In any case, you should get expressions involving $\binom{2k}{k}$, where you can apply Stirling's formula (for sufficiently large $k$) and get the result.
Hope this can get you started.
