If $D(m)$ is the deficiency of the deficient number $m$, then what is $\lim_{m \rightarrow \infty}{\frac{D(m)}{m}}$? Let $\sigma=\sigma_1$ denote the classical sum-of-divisors function. Call the function $D(x)=2x-\sigma(x)$ as the deficiency of $x$.
Here is my question:

If $D(m)$ is the deficiency of the deficient number $m$, then what is $$\lim_{m \rightarrow \infty}{\frac{D(m)}{m}}?$$

 A: Hint: Both primes and powers of two are deficient, and the limit of these subseqeunces are $1$ and $0$, respectively.
A: The limit does not exist. I can take arbitrarily large primes, which have $D(p) = 2p - (1+p) = p-1$, and thus have $\frac{D(p)}{p} = 1+\frac{1}{p-1} \to 1$. I could also consider $q=2p$, which have $D(2p) = 4p - (1+2+p+2p) = p - 3, \implies \frac{D(q)}{q} = \frac{p-3}{2p} \to \frac{1}{2}$. Since this function attains values arbitrarily closed to $\frac{1}{2}$, and numbers arbitrarily close to 1, infinitely often, it does not converge.
A: Preamble:  This is for the OP's own benefit.  I leave it here for the sake of posting a copy of my own thoughts on the problem.
Note that
$$\frac{D(m)}{m} = \frac{2m-\sigma(m)}{m} = 2-\frac{\sigma(m)}{m} = 2-I(m)$$
where
$$I(m)=\frac{\sigma(m)}{m}$$
is the abundancy index of $m$.
Furthermore, we have
$$\prod_{i=1}^{\omega(m)}{\left(\frac{p_i + 1}{p_i}\right)} \leq I(m) < \prod_{i=1}^{\omega(m)}{\left(\frac{p_i}{p_i - 1}\right)}$$
given that $m$ has $\omega(m)$ prime factors $p_1 < p_2 < \ldots < p_{\omega(m)}$.
Thus, if
$$\frac{D(m)}{m}$$
were to have a "limit", then it would be $2$ minus the "limit" of $I(m)$.
But the "limit" of $I(m)$ is obtained by letting at least one prime $p_i \rightarrow \infty$, which then gives
$$1 = \lim_{p_i \rightarrow \infty}{\prod_{i=1}^{\omega(m)}{\left(\frac{p_i + 1}{p_i}\right)}} \leq \lim_{m \rightarrow \infty}{I(m)} \leq \lim_{p_i \rightarrow \infty}{\prod_{i=1}^{\omega(m)}{\left(\frac{p_i}{p_i - 1}\right)}} = 1,$$
so that
$$\lim_{m \rightarrow \infty}{\frac{D(m)}{m}} = \lim_{m \rightarrow \infty}{\left(2-I(m)\right)} = 2 - 1 = 1.$$
Update: August 27 2016
It turns out that there are convergence issues (i.e. with subsequences) that must be considered.  Hence, this "proof" is not valid.
