Trying to understand proof that price of anarchy O(logm/loglogm) for m links Let $t \geq \dfrac{3\ln m}{\ln\ln m}$ where $m$ is a positive integer.
I'm trying to see why:
$$t + \sum_{\tau = t}^\infty m\cdot\left(\dfrac{2e}{\tau}\right)^\tau \leq t + \sum_{\tau = t}^\infty 2^{-\tau} \leq t + 1$$
I don't see it. Also, do we need the $m$ for this equation to be true?
http://www.dcs.warwick.ac.uk/~czumaj/PUBLICATIONS/DRAFTS/SODA-2002-Full.pdf
See right above theorem 4
 A: This is really not an answer but I hope someone can use my work to come-up with one.
I don't know how to compute $\sum\limits_{\tau=t}^{+\infty}\left(\frac{2e}{\tau}\right)^\tau$ so my idea was to compare the two series term by term:
$$ m\cdot \left(\frac{2e}{\tau}\right)^\tau \le 2^{-\tau} \Leftrightarrow m\cdot \left(\frac{4e}{\tau}\right)^\tau \le 1 \Leftrightarrow \ln(m)+\tau\cdot\left(\ln(4)+1-\ln(\tau)\right)\le 0.\tag{1}$$
We also have the condition: $$\tau\ge t \ge \frac{3\ln(m)}{\ln(\ln(m))}.\tag{2}$$
If we can prove that (2) implies (1) for all $m$, we would have answered the question.
Now $\tau\cdot\left(\ln(4)+1-\ln(\tau)\right)$ increases before its unique maximum at $\tau=4$ and  decreases afterwards. On the other hand: $\frac{3\ln(m)}{\ln(\ln(m))}$ has its unique minimum, which is $3e$, at $e^e$. We can conclude that $\tau\cdot\left(\ln(4)+1-\ln(\tau)\right)$ is always decreasing in $\tau$ in the range of interest as $\tau>3e>4$. 
The practical consequence of the above is that now it suffices to prove that (1) holds for $\tau=\frac{3\ln(m)}{\ln(\ln(m))}$. But the plot of $\ln(m)+\frac{3\ln(m)}{\ln(\ln(m))}\left(\ln(4)+1-\ln\left(\frac{3\ln(m)}{\ln(\ln(m))}\right)\right)$ lies below zero only for $m\le 4$. 
Assuming I didn't do mistakes, the above shows that comparing terms by terms only works for $m\le 4$.
