It seems from your notation and the second part of your question that you are summing over the prime divisors. In that case there is no need to mention $k$ and one can get arbitrarily close to 1. There is an infinite sequence $2,3,5,7,11,29,127,1931,309121,47777896349 \dots$ where each term is the smallest prime not already listed which keeps $\sum \frac{1}{p+1}$ strictly less than 1. The sum for $2,3,5,7,11,23$ is exactly 1. Curiously, the partial sums starting with $\frac{3}{4}$ have the form $\frac{t-1}{t}$ until about the 11th term.

Your the lead in of your second question could be reworked into: 

Pick a prime $p=p_1$ and consider  $\sum_{i=1}^{n}\frac{1}{p_i+1}$ where the $p_i$ are consecutive primes and $n=n(p)$  is the largest integer which keeps the sum under 1.

 $n(2)=5,n(3)=13,n(5)=37,n(7)=84,n(11)=171$ (The 171 primes from 11 to 1039 have $\sum \frac{1}{p+1}$ about $1-\frac{1}{5261}$. The 290 primes from 13 to 1933 have $\sum \frac{1}{p+1}$ about $1-\frac{1}{10179}$). Based on very limited evidence, it seems possible that $1-\sum_{i=1}^{n}\frac{1}{p_i+1}>\frac{1}{n^2}$ for $n>n(7)=84$. However I certainly can't rule out it being freakishly close to 1.