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A telescoping series has partial sums in which internal contributions cancel. The basic pattern is $a_n=b_n-b_{n+1}$, for which finite summation givesOnly the two boundary terms survive. If $b_n$ tends to a finite limit of a sequence $L$, the series consequently sums to $b_1-L$. The cancellation is exact for each finite $N$; the infinite result then follows by a limit of a sequence.

For example, partial fraction decomposition givesThe first three terms are $1/2,1/6,1/12$, with partial sum $3/4$. The first five have partial sum $5/6$. Taking the limit of a sequence gives the concise answer

The appeal is that a long calculation becomes a boundary calculation. It works whenever the summand is a successive difference, not only for rational functions. For instance, $a_n=\log ((n+1)/n)=\log (n+1)-\log n$ gives partial sum $\log (N+1)$, so its series diverges despite the cancellation. A telescoping series therefore need not be a convergent series: one must still examine the surviving boundary term.

The term test for divergence says that a series cannot converge if its terms fail to tend to zero. Indeed, if its partial sums satisfy $S_n\to S$, then $a_n=S_n-S_{n-1}\to S-S=0$. This is a necessary condition for a convergent series.

The converse fails: $1/n\to 0$, but the harmonic series diverges. For a quick successful test, ${\sum }_{n=1}^{\infty }n/(n+1)$ diverges because its terms tend to one. The test gives no conclusion for $\sum 1/n$ or $\sum 1/[n(n+1)]$: both have terms tending to zero, yet only the second is a convergent series. The telescoping series calculation and the harmonic series block estimate resolve those cases.

A variable-coefficient affine recurrence is a linear recurrence relation with coefficients allowed to depend on the index. If every $a_n$ is nonzero, a discrete integrating factor reduces it to a telescoping series.