Proof of bound for Lovasz Local lemma I'm trying to prove, using a probabilistic argument, that the list coloring number of a bipartite graph of degree at most $d$ is $O\left(\frac{d}{log(d)}\right)$.
For this, I have built a Lovasz graph and in order be able to apply the Lovasz local lemma I need to show that
$$ e (d^2-d+1) \left[1 - \left(1-\frac{1}{t}\right)^d\right]^t \leq 1$$
if $t = O\left(\frac{d}{log(d)}\right)$
I've been trying to work out the math for a while now but so far it hasn't taken me anywhere. Any suggestions as to how to proceed?
 A: Take $t=\frac{kd}{\log{d}}$ and we have
$$
\begin{align*}
e(d^2-d+1)\left(1-\left(1-\frac{1}{t}\right)^d\right)^t &= e^{\log(d^2-d+1)+1}\left(1-\left(1-\frac{1}{\frac{kd}{\log{d}}}\right)^d\right)^{\frac{kd}{\log{d}}}\\
&= e^{\log(d^2-d+1)+1}\left(1-\left(1-\frac{1}{\frac{kd}{\log{d}}}\right)^{\frac{kd}{\log{d}}\cdot\frac{\log{d}}{k}}\right)^{\frac{kd}{\log{d}}}\\
&\approx e^{\log(d^2-d+1)+1}\left(1-e^{-\frac{\log{d}}{k}}\right)^{\frac{kd}{\log{d}}}\\
&= e^{\log(d^2-d+1)+1}\left(1-\frac{1}{d^{\frac{1}{k}}}\right)^{\frac{kd}{\log{d}}}\\  
&= e^{\log(d^2-d+1)+1}\left(1-\frac{1}{d^{\frac{1}{k}}}\right)^{d^{\frac{1}{k}}\cdot \frac{kd^{(k-1)/k}}{\log{d}}}\\ 
&\approx e^{\log(d^2-d+1)+1}e^{\frac{-kd}{\log{d}}}\\
\end{align*}
$$
For $k$ sufficiently large.
Then we want the exponent to be less than $0$ in order for the entire expression to be smaller than $1$. That means
$$
\log{d}(\log(d^2-d+1)+1)-kd\le0\iff k\ge\frac{\log{d}(\log(d^2-d+1)+1)}{d}.
$$
Now
$$
\frac{\log{d}(\log(d^2-d+1)+1)}{d}\le\frac{\log{d}(\log(d^2)+1)}{d}\le\frac{2(\log{d})^2}{d}+\frac{1}{e}=f(d)
$$
and $f(d)$ attains a maximum at $d=e^2$ so that $f(e^2)=\frac{8}{e^2}+\frac 1e\le 1.46.$ Thus for $k$ sufficiently large, our claim holds.
