I need to find out an estimation or any bounds for what $\log(k)$ is when $k = \lceil \frac{n}{2\log(n)} \rceil$.

I have reached a point in my calculations where I am left with $\left( \frac{n - \frac{1}{2}}{\log(k)} + 1 \right)$ and I need that to somehow relate to $(1 + \epsilon) \frac{n}{\log(n)}$.

But I can't seem to find an approximation for $\log\left( \frac{1}{\log(n)} \right)$.

Any help will be useful as I desperately need an answer asap, thanks!!

  • 2
    $\begingroup$ it's almost as dangerous as dividing by zero, proceed with caution, and under your own risk. $\endgroup$
    – Asinomás
    Aug 28 '16 at 18:58
  • 1
    $\begingroup$ Well, $\log ( 1/\log n) = -\log \log n$ and $\log log (\mbox{googol})$ is only about 5.5. Context might help here. Are you analyzing an algorithm? $\endgroup$
    – B. Goddard
    Aug 28 '16 at 22:00
  • $\begingroup$ More background is given below in the answer section @B.Goddard $\endgroup$
    – esther21
    Aug 29 '16 at 22:31

I am analysing a Maker-Breaker game on $n$ vertices where,

k = $\lceil{\frac{n}{2\log{n}}}\rceil$ and $b$ $>$ ($1+$$\epsilon$)$\frac{n}{\log{n}}$

I am left with this equation

\begin{equation} k(n-k) \leq (b-1)k \sum_{i=1}^{k-1} \frac{1}{i} \end{equation}

After I tried to substitute and simplify I can only get up to the stage I posted before

that $b > \frac{n-\frac{1}{2}}{\log{k}} +1 $

and thats where I am stuck. Please let me know if you notice anything I have done wrong


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