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Since the question has acquired over 1000 views (of which I am sure less than half were done by me), I thought I would celebrate by giving an improvement that I promised half a year ago. This is a refinement of a version I sent to several people; I invite the reader to submit corrections and/or criticism.


I will find $T'$ big enough that $T' > T$, yet small enough to get a nice value for $s$ and show that $P - T' > 0$. Then I will massage $P - T'$ and $SB$ to give a slightly weaker upper bound that is easier to write down.

Let's write $\sigim = \sum_{1<=i<=n} 1/m_i$ with $m_i$ being the distinct prime divisors of $m$. (I use $-1$ as a superscript, NOT as an exponent, in both $\piim$ and $\sigim$.) Now $T$ is an alternating series, and for $k \gt 0$, $$\sigim\sum_{d \mid m, \nu(d)=k} 1/d \gt (k+1)\sum_{d \mid m, \nu(d)=(k+1)} 1/d,$$ so if $s$ is odd and larger than $\sigim$, then it would suffice to replace $T$ by $D=\sum_{d \mid m, \nu(d)=s+1} 1/d$, provided we can show $D < P$. Instead, we use an upper bound for $D$, namely $T' = \sigim^{s+1}/(s+1)!$, which follows by using the inequality above $s$-many times.

Let us find a value for $s$ such that $P(s+1)! > \sigim^{s+1} = (s+1)!T'$. An earlier version of this result did some work to show that one could pick $s+1 >= 4\sigim > 0$, and in fact $4$ can sometimes be replaced by a smaller constant. The choice of $T'$ saves some work, and using $4$ will also make things easier.

The following steps require $m \gt 1$, $s+1 \gt 1$, $s+1 \geq 4\sigim$, $0\lt P \lt 1$, and finally $e \lt P^{-1/\sigim} \leq 4 \in (e,4]$ (which is proved in [1]).

\begin{eqnarray*} & e \lt 4^{3/4} \text{, so } e* P^{-1/4\sigim}\lt e*4^{1/4} \lt 4, \\ \text{so} & \sigim \lt P^{1/4\sigim} 4\sigim/e \leq P^{1/s+1}((s+1)/e), \\ \text{so} & \sigim^{s+1} \lt P((s+1)/e)^{s+1} \lt P(s+1)! . \end{eqnarray*}

Now that we have a candidate for $s$, let us choose the smallest odd $s$ such that $s+1 \geq 4\sigim$. Then


where this is most useful for $1 \le \sigim$ and $s$ the smallest odd integer greater tham $4\sigim$.

I may show some later refinements of this. However, $\sigim < 1 + \log\log n$ for $n > 7$ so the exponent $s$ grows very slowly. This will do for now.

[1] G. Paseman, "The Waltraud and Richard R. Paseman Theorem", private manuscript, March 2011.