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, that partial sums starting with $\frac{3}{4}$ have the form $\frac{n-1}{n}$ until about the 11th term.

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.