Partial fraction telescoping ($\frac{1}{(3k-2)(3k+1)}=\frac{1}{3}\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)$)

Sometimes a sum does not initially look like a difference of neighboring entries. A partial fraction decomposition can reveal the cancellation. For the original examplesubtract the two reciprocals:ThusFor $N=3$, direct addition gives $1/4+1/28+1/70=3/10$. For $N=6$, the six positive entries add to $6/19\approx 0.315789$. The partial sums approach $1/3$, and the exact series remainder isThis formula answers an accuracy question as well as a summation question. To make the remaining positive series remainder smaller than $0.001$, solve $1/[3(3N+1)]<1/1000$. The result is $9N+3>1000$, so the smallest permitted integer is $N=111$. At $N=110$, the error is $1/993>0.001$; at $N=111$, it is $1/1002<0.001$.