Looking for suggestions on how to proceed with showing that:

Question

for $x \ge 2863:$

$$\ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) < \sum_{k=5}^{\infty}-\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right)$$

I've written a java application which checked from 2,863 to 24,600 which is how I came up with $2,863$.

I'm looking for tips on how to proceed to prove or disprove this comparison.

Since:

$$\ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) > \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

The problems comes down to showing that for $x \ge 2863$:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le 0$$

or showing that:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le \ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) - \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

Does anyone have any suggestions?

Thanks very much!

-Larry

Answer 1

Did you mean $$ \sum_{k=6}^\infty \mu(k) \ln\left(\left\lfloor \frac{x}{k} \right\rfloor !\right) \geq 0 $$

If not, I can't reproduce your verification for this inequality using the following Matlab code:

function s = larry( x )

    % Compute \sum_{k=6}^\infty \mu(k) \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    s = 0;
    for k = 6:ceil(x/2)
        s = s + moebiusmu(k)*logfactorial( floor(x/k) );
    end

    % Compute \mu(k)
    function mu = moebiusmu(n)

        if n == 1, mu = 1; return; end

        p = factor(n); 
        r = histc(p, unique(p));

        if all(r < 2) 
            mu = (-1)^length(unique(p)); 
        else 
            mu = 0; 
        end
    end

    % Compute \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    function p = logfactorial(n)        
        if n < 2
            p = 0;
        else
            p = sum(log( 2:n ));
        end
    end

end