Originally Posted by

**Evo Rob** Firstly, please accept my humble apologies, there was a mistake in my typing of the problem

formulation. For some inexplicable reason, I typed in ceiling functions when they should have been

floors. This is possibly what caused the discrepancy in the first few values of f(k)

However, the suggested approach works, with a few minor alterations, for the correct problem

formulation:

$\displaystyle

f(k) ={\frac{1}{\Omega}}\sum_{j=k}^{\lfloor {k/\Omega}

\rfloor}{\frac{1}{j}}-{\frac{\lfloor{\frac{k}{\Omega}}

\rfloor}{k}}+{\frac{1}{\Omega}}+{\frac{2}{k}}-{\frac{2}{k\Omega}}

$

As $\displaystyle 1/x$ is a positive decreasing function for $\displaystyle k>1$, we have:

$\displaystyle

\sum_{j=k}^{\lfloor(k/\Omega\rfloor} \frac{1}{j}\le \int_{x=k}^{\frac{k}{\Omega}} \frac{1}{x-1}\

dx=\ln(k/\Omega -1) - \ln(k-1)=\ln\left( \frac{k/\Omega -1}{(k-1)}\right)

$

Also, as

$\displaystyle

\lfloor{k/\Omega}\rfloor\ge{k/\Omega}-1

$

we have:

$\displaystyle

f(k)\le {\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}

$

Now let:

$\displaystyle

h(k)={\frac{1}{\Omega}}\ln\left( \frac{k/\Omega -1}{(k-1)}\right) + {\frac{3}{k}}-{\frac{2}{k\Omega}}

$

$\displaystyle

{\frac{dh}{dk}}=-{\frac{(1-\Omega)}{(k-1)(k-\Omega)}}- {\frac{3}{k^2}}+{\frac{2}{k^2\Omega}}

$

Now for $\displaystyle {\frac{dh}{dk}} $ to be positive, requires that:

$\displaystyle

{\frac{(k-1)(k-\Omega)}{k^2}}>{\frac{\Omega(1-\Omega)}{(2-3\Omega)}} = 0.8222248$

This is true $\displaystyle \forall k\ge9 $.

As $\displaystyle \lim_{k\to \infty}h(k)=1$ and $\displaystyle

\frac{dh}{dk}>0\ \forall k>8

$

$\displaystyle

f(k)\le h(k) \le1\ \forall k>8

$

Finally, $\displaystyle f(k)$ can be evaluted for $\displaystyle k=1$ to $\displaystyle 8$, completing the proof.

Once again, many thanks for the excellent and timely help with this, it is much appreciated.

Rob