= Telescoping series
{title2=$\sum_{n=1}^N(b_n-b_{n+1})=b_1-b_{N+1}$}
{wiki}
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> gives
$$
\sum_{n=1}^{N}(b_n-b_{n+1})
=(b_1-b_2)+(b_2-b_3)+\cdots+(b_N-b_{N+1})
=b_1-b_{N+1}.
$$
Only 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> gives
$$
\frac1{n(n+1)}=\frac1n-\frac1{n+1},\qquad
\sum_{n=1}^{N}\frac1{n(n+1)}=1-\frac1{N+1}.
$$
The 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
$$
\boxed{\sum_{n=1}^{\infty}\frac1{n(n+1)}=1}.
$$
\Image[/telescoping-cancellation.png]
{title=Five reciprocal-product terms cancel internally, leaving only the first and last boundary contributions}
{description=Matching colors pair the negative $1/(n+1)$ in one term with the positive $1/(n+1)$ in the next. The cancellation leaves $1-1/6=5/6$, the fifth <partial sum>, rather than the full infinite sum.}
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.
Back to article page
