Why Sum $> 1/2$ in proving reciprocal of prime diverges In my number theory book it says that to show that the sum of the reciprocals of the primes diverges, it’s enough to show that, for any $j$:
$$\frac{1}{p_{j+1}}+\frac{1}{p_{j+2}}+\frac{1}{p_{j+3}}+\cdots>\frac{1}{2}$$
Why is it enough to show that the above sum is greater than $1/2$? 
 A: If $\sum \frac{1}{p_i} = N$ converges, then by definition
\begin{align*}
\left|N - \sum_{i \leq n} \frac{1}{p_i}\right| = \sum_{i > n} \frac{1}{p_i} \to 0
\end{align*}
as $n\to\infty$.
A: Taking "any" to mean "every", the point would be that no matter how large $j$ gets, that sum remains bigger than $1/2$.  If the sum of the reciprocals converges, then after some finite number of terms, the partial sum is more than the total sum minus $0.001$.  Take $j$ bigger than that number of terms, and the tail of the series would then add up to less than $0.001$.  And similarly for all other small positive numbers.  But if the tail never gets less than $1/2$, then it can't get less than $0.001$ or any of those other tiny numbers.
To put it another way, if the series $\displaystyle\frac 1 {p_1}+\frac 1 {p_2}+ \frac 1 {p_3} + \cdots$ converges, then
$$
\frac 1 {p_j} + \frac 1 {p_{j+1}} + \frac 1 {p_{j+2}} + \cdots \to 0\text{ as }j\to\infty.
$$
A: The idea is that if the sum converges, after a bunch of terms it is close to the final sum.  This seems to be the concept you are missing. As all the terms are positive, proving the tail exceeds $\frac 12$ or any other positive number proves that the sum is not close to any proposed limit.  In this case a slightly more formal proof might help.  To prove a sum converges to $L$, we need to prove that for any $\epsilon \gt 0$ there is an $N$ so that the sum of at least $N$ terms is within $\epsilon$ of $L$.  If the tail sum is always greater than $\frac 12$ and I give you $\epsilon = \frac 14$, you can't find an $N$ that works.
