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For a real sequence $(a_n)$, the partial sum of ${\sum }_{k=1}^{\infty }(a_k-a_{k-1})$ is exactly $a_n-a_0$. Consequently it is a convergent series if and only if $a_n$ has a finite limit $L$, and its sum is $L-a_0$. Cancellation takes place inside each finite partial sum; taking a limit is a separate step.^[ref]

The partial fraction telescoping example has $S_N=N/(3N+1)$ and therefore limit $1/3$. In contrast, let $a_n=(-1{)}^n$. Its first entry is $a_0=1$, so $S_N=(-1{)}^N-1$. The even partial sums are zero and the odd partial sums are negative two. Both finite cancellations are correct, but the sequence has no single limit, giving a divergent series.

It is also essential to keep the two contributions paired. For example, ${\sum }_{k=1}^{\infty }(1/k-1/(k+1))$ is a convergent series with sum one, as its finite partial sum is $1-1/(N+1)$. Writing it instead as ${\sum }_{k=1}^{\infty }1/k-{\sum }_{k=1}^{\infty }1/(k+1)$ asks to subtract two divergent series. The permitted finite cancellation does not define that separate subtraction. Working through partial sums keeps the indices aligned and the operation meaningful.

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 $1/1002<0.001$.

For a convergent series with sum $S$ and $N$-th partial sum $S_N$, its series remainder is $R_N=S-S_N$. It equals the sum of the terms after position $N$. Its absolute value measures the error of replacing $S$ by $S_N$; it need not be positive when the series has signed sequence terms.