Is the sum of the reciprocals of primes none of whose proper initial segments are primes converge? Is the sum of the reciprocals of primes none of whose proper initial segments are primes converge?
An example of those primes is $88547$, because $8,88,885$ and $8854$ are not primes. 
The first few terms in the sequence (A069090) are:
   2, 3, 5, 7, 11, 13, 17, 19, 41, 43, 47, 61, 67, 83, 89, 97, 101, 103, 107, 109, 127, 149, 151, 157, 163, 167, 181, 401, 409, 421, 443, 449, 457, 461, 463, 467, 487, 491, 499, 601, 607, 631, 641, 643, 647, 653, 659, 661, 683, 691, 809, 811, 821, 823, 827

 A: The series converges; the limit is approximately $2.48111396244$.
That the series converges follows from the Kraft–McMillan inequality: A uniquely decodable code, and hence in particular a prefix code, satisfies
$$
\sum_ir^{-l_i}\le1\;,
$$
where $r$ is the size of the alphabet and $l_i$ is the length of the $i$-th codeword. In the present case, the primes without proper prime prefix form a prefix code $\{q_i\}$ with $r=10$ and $l_i-1\lt\log_{10}q_i$, so
$$
\sum_i\frac1{q_i}=\sum_i10^{-\log_{10}q_i}\le\sum_i10^{-(l_i-1)}\le10\;.
$$
The series converges very slowly when summed directly. To compute the limit efficiently, consider the evolution of the set $S\subset[1,10)$ of reals that do not have one of the primes included in the sum so far as a prefix (in decimal notation, without the decimal point). For instance, after excluding $2, 3, 5, 7$ as prefixes, we are left with $S=[1,2)\cup[4,5)\cup[6,7)\cup[8,10)$, and upon excluding  the prefix $11$ this is reduced to $S=[1,1.1)\cup[1.2,2)\cup[4,5)\cup[6,7)\cup[8,10)$.
In each range of numbers $\left[10^n,10^{n+1}\right)$ approximately a fraction $\frac1{\log10^n}=\frac1{n\log10}$ of numbers are primes, so on the random model of the primes each such range reduces the measure of $S$ by approximately that fraction. For large $n\gt n_0$, we can approximate the measure $\mu_n$ left in $S$ by integrating the logarithms of the corresponding factors:
\begin{eqnarray}
\log\mu_n
&\approx&
\log\mu_{n_0}+\int_{n_0}^n\log\left(1-\frac1{k\log10}\right)\mathrm dk
\\
&\approx&
\log\mu_{n_0}-\int_{n_0}^n\frac1{k\log10}\mathrm dk
\\
&=&
\log\mu_{n_0}-\frac{\log\frac n{n_0}}{\log 10}\;,
\end{eqnarray}
so
$$
\mu_n\approx\mu_{n_0}\left(\frac n{n_0}\right)^{-\frac1{\log10}}\;.
$$
Thus, the measure of $S$ tends to zero. That means that except for a set of measure $0$, every $x\in [1,10)$ has a prime prefix in decimal notation.
This allows us to compute the desired limit far more efficiently than by directly summing the reciprocals. If a prime $p\in\left[10^n,10^{n+1}\right]$ were to contribute
$$
\int_{\frac p{10^n}}^{\frac{p+1}{10^n}}\frac{\mathrm dx}x=\log\left(1+\frac1p\right)
$$
to the sum, then, since all of $[1,10)$ is eventually covered, the limit of the sum would be
$$
\sideset{}{'}\sum_p\log\left(1+\frac1p\right)=\int_1^{10}\frac{\mathrm dx}x=\log10
$$
(where the prime indicates that the sum only runs over the primes without proper prime prefix). Instead, $p$ contributes $\frac1p$, which differs from $\log\left(1+\frac1p\right)$ only on the order of $p^{-2}$. Thus, we have
\begin{eqnarray}
\sideset{}{'}\sum_p\frac1p
&=&
\sideset{}{'}\sum_p\frac1p+\log10-\sideset{}{'}\sum_p\log\left(1+\frac1p\right)
\\
&=&
\log10+\sideset{}{'}\sum_p\left(\frac1p-\log\left(1+\frac1p\right)\right)\;.
\end{eqnarray}
(Note that we're not relying on the specific form $\mu_n\sim n^{-1/\log10}$ estimated above, only on $\lim_{n\to\infty}\mu_n=0$.)
Here's Java code that finds the primes without proper prime prefix and sums $\frac1p$ and $\frac1p-\log\left(1+\frac1p\right)$. The sum of $\frac1p-\log\left(1+\frac1p\right)$ converges to double machine precision with $p\le10^8$; the result (plus $\log10$) is $2.48111396244$. By contrast, with $p\le10^8$ the direct summation of $\frac1p$ only reaches $2.01$. With $p\le10^{12}$, it’s still at $2.09$. Below is a plot of the partial sums of $\frac1p$ for $p\le x$. The abscissa is $\log\log x$; the ordinate is $-\log(S-S(x))$, where $S(x)$ is the partial sum and $S$ is the limit (as determined from the rapidly converging sum). Since the fraction of the sum that's missing at $p\le x=10^n$ is asymptotically proportional to $n^{-1/\log10}\propto(\log x)^{-1/\log10}$, we'd expect this plot to be asymptotic to a line with slope $-\frac1{\log10}$. The green line is a line with that slope fitted to the data. The good agreement of the fit with the data confirms the value of the limit.

