Telescoping series (${\sum }_{n=1}^N(b_n-b_{n+1})=b_1-b_{N+1}$)

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.

For the telescoping series ${\sum }_{n=1}^{\infty }1/[n(n+1)]=1$, the remainder after the first $N$ terms is exactlyThus $S_9=0.9$, $S_{99}=0.99$, and $S_{999}=0.999$. This supplies an exact stopping rule: to make the absolute error at most $10^{-3}$, 999 terms suffice. Requiring strictly less than $10^{-3}$ instead needs at least 1000 terms.

The next term, $a_{N+1}=1/[(N+1)(N+2)]$, is smaller than the full remaining sum by a factor of $N+2$. For example, after 99 terms the next term is $1/10100$, while the remainder is $1/100$. The term test for divergence concerns individual terms; a reliable numerical approximation needs a bound on the whole remaining sum.