telescoping

Telescoping trick

If consecutive terms of a series cancel each other, most of the sum collapses and only the first and last terms survive.

If $a_k=b_k-b_{k+1}$, then:

$$\begin{array}{rl}\sum _{k=1}^n{a}_k& =\sum _{k=1}^n(b_k-b_{k+1})\\ & =(b_1-b_2)+(b_2-b_3)+(b_3-b_4)+\cdots +(b_n-b_{n+1})\\ & =b_1-b_{n+1}\end{array}$$

Everything in the middle cancels pairwise.

What is the form of a telescoping sum?

$$\sum _{k=m}^n{a}_k=\sum _{k=m}^n(b_k-b_{k+1})$$

What does ${\sum }_{k=m}^n(b_k-b_{k+1})$ simplify to?

$b_m-b_{n+1}$

${\sum }_{k=1}^{\infty }\frac{1}{k(k+1)}=1$

Write $1=(k+1)-k$ in the numerator and split:

$$\frac{1}{k(k+1)}=\frac{(k+1)-k}{k(k+1)}=\frac{\cancel{k+1}}{k\cancel{(k+1)}}-\frac{\cancel{k}}{\cancel{k}(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$

So $b_k=\frac{1}{k}$ and:

$$\sum _{k=1}^n\frac{1}{k(k+1)}=\frac{1}{1}-\frac{1}{n+1}\stackrel{n\to \infty }{\to }1$$

Telescoping is the discrete fundamental theorem of calculus

With the finite difference $\Delta b_k=b_{k+1}-b_k$,

$$\sum _{k=1}^n\Delta b_k=b_{n+1}-b_1$$

This is the discrete analog of ${\int }_a^b{F}^{\prime }(x)dx=F(b)-F(a)$ (fundamental theorem of calculus): summing a difference recovers the net change between the endpoints. Writing $a_k=b_k-b_{k+1}$ is finding a discrete antiderivative of $a_k$.

Note

If $a_k=b_k-b_{k+\ell }$ for some fixed $\ell \in \mathbb{N}$, then the sum telescopes with $\ell$ surviving terms at each end:

$$\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\ell }{b}_k-\ell \cdot \underset{k\to \infty }{\lim }{b}_k$$

(assuming the limit exists)

Note: $b_{k+l}=b_{k+1}+b_{k+2}+\dots b_{k+l}$, so ${\lim }_{k\to \infty }{b}_{k+l}={\lim }_{k\to \infty }(b_{k+1}+b_{k+2}+\dots b_{k+l})=l\cdot {\lim }_{k\to \infty }{b}_k$

What does the ${\sum }_{k=1}^{\infty }{b}_k-b_{k+l}$ simplify to?

$$\begin{array}{rl}{S}_N& :=\sum _{k=1}^N(b_k-b_{k+l})\\ & =\sum _{k=1}^N{b}_k-\sum _{k=1}^N{b}_{k+1}\\ & =\sum _{k=1}^N{b}_k-\sum _{k=l+1}^{N+l}{b}_k\\ & =\sum _{k=1}^l{b}_k-\sum _{k=N+1}^{N+l}{b}_k\\ & \stackrel{N\to \infty }{\to }\sum _{k=1}^l{b}_k-l\cdot \underset{k\to \infty }{\lim }{b}_k\end{array}$$

Define the partial sum $S_N$, split the sum, shift the index, so the first $l$ and last $l$ terms survive, the rest cancels. Then take the limit as $N\to \infty$. It’s just $l$ times the limit of $b_k$ because $k\to \infty$.

Recognising telescoping

Telescoping applies when the summand is a difference $a_k=b_k-b_{k+\ell }$, so terms $\ell$ apart cancel and only the ends survive. Two cases let you put it in that form.

The summand is already a difference of some $f$ at shifted arguments: $\ln \frac{k+1}{k}=\ln (k+1)-\ln k$, or $\sqrt{k+1}-\sqrt{k}$, giving $b_k=f(k)$.

The summand is rational, and partial fractions produce the difference: $\frac{1}{k(k+1)}$, $\frac{1}{(2k-1)(2k+1)}$.
The offset between the denominator’s factors is the lag $\ell$ at which terms cancel: $k$ and $k+1$ cancel against the next term ($\ell =1$); $k$ and $k+2$ cancel two terms later ($\ell =2$, two survivors at each end).
Shortcut: $\frac{1}{k(k+m)}=\frac{1}{m}\left(\frac{1}{k}-\frac{1}{k+m}\right)$.

If a problem asks for the exact sum of a non-geometric series, it was likely built to telescope.

telescoping product

$$\prod _{k=m}^n\frac{b_{k+1}}{b_k}=\frac{b_{n+1}}{b_m}$$

Consecutive factors cancel, only the ends survive.

Link to original