Telescope sum

Pushing together a telescope - namesake of the telescope sum

A collapsible telescope

A telescoping series is in the mathematics a finite sum of differences (except the first and the last) cancel each other, wherein each two neighboring members. This process is called telescoping a sum . The term is derived from pushing two or more cylindrical tubes into one another .

If there is a consequence , then is ${\displaystyle (a_{k})_{k\in \mathbb {N} }\,}$

${\displaystyle \sum _{i=1}^{n}(a_{i}-a_{i+1})}$

a telescope sum. If you can write a sum as a telescope sum, its evaluation is simplified:

${\displaystyle \sum _{i=1}^{n}(a_{i}-a_{i+1})=(a_{1}-a_{2})+(a_{2}-a_{3})+\cdots +(a_{n-1}-a_{n})+(a_{n}-a_{n+1})=a_{1}-a_{n+1}}$.

A series whose partial sums are telescope sums is called a telescope series . A row of telescopes

${\displaystyle \sum _{i=1}^{\infty }(a_{i}-a_{i+1})}$

is convergent if and only if converges to a limit value . The sum of the series is then the same . ${\displaystyle (a_{k})_{k\in \mathbb {N} }\,}$ ${\displaystyle g\,}$ ${\displaystyle a_{1}-g\,}$

The same applies to a product:

${\displaystyle \prod _{i=1}^{n}{\frac {a_{i+1}}{a_{i}}}={\frac {a_{2}}{a_{1}}}\cdot {\frac {a_{3}}{a_{2}}}\cdot {\frac {a_{4}}{a_{3}}}\cdots {\frac {a_{n}}{a_{n-1}}}\cdot {\frac {a_{n+1}}{a_{n}}}={\frac {a_{n+1}}{a_{1}}}}$

is a telescope product , so to speak .

The situation is more complicated when the telescope runs over three (or more) consecutive links (see example).

In number theory telescopic sums represent an important tool.

Examples

• finite geometric series :

${\displaystyle (x-1)\sum _{k=0}^{n}x^{k}=\sum _{k=0}^{n}(x^{k+1}-x^{k})=x^{n+1}-1.}$

• Telescope sums are often a bit hidden and can be recognized, for example, by partial fraction decomposition :

The partial fraction decomposition of can be obtained using, for example ${\displaystyle {\frac {1}{k(k+1)}}}$

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

It follows

${\displaystyle \sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)=1-{\frac {1}{n+1}}.}$

• three times the telescope sum:

${\displaystyle {\begin{aligned}(x-1)^{2}\sum _{k=1}^{n}kx^{k-1}&=\sum _{k=1}^{n}(kx^{k+1}-2kx^{k}+kx^{k-1})=\sum _{k=1}^{n}[kx^{k+1}-(k-1)x^{k}]-\sum _{k=1}^{n}[(k+1)x^{k}-kx^{k-1}]\\&=nx^{n+1}-(n+1)x^{n}+1.\end{aligned}}}$

Alternatively, this follows for by differentiation from the first example with the help of the quotient rule : ${\displaystyle x\neq 1}$

${\displaystyle \sum _{k=1}^{n}kx^{k-1}={\frac {\rm {d}}{{\rm {d}}x}}\sum _{k=0}^{n}x^{k}={\frac {\rm {d}}{{\rm {d}}x}}{\frac {x^{n+1}-1}{x-1}}={\frac {(n+1)x^{n}(x-1)-(x^{n+1}-1)}{(x-1)^{2}}}={\frac {nx^{n+1}-(n+1)x^{n}+1}{(x-1)^{2}}}}$.

This is an important application example of differential calculus as a calculus for term transformation .

literature

• Rolf Walter: Introduction to Analysis. Volume 1. Walter de Gruyter, Berlin et al. 2007, ISBN 978-3-11-019539-2 , pp. 31–32 ( excerpt in the Google book search).
• Harro Heuser : Textbook of Analysis. Volume 1. 6th edition, unchanged reprint of the 5th revised edition. Teubner, Stuttgart et al. 1989, ISBN 3-519-42221-2 , pp. 91, 94, 194.

Web links

Wikibooks: Math for Non-Freaks: Telescope Sum and Telescope Series  - Learning and Teaching Materials

• Telescoping Sum on PlanetMath
• Po-Shen Loh: Telescoping Sums and Products (PDF; 66 kB) - examples of telescope sums and telescope products
• Philippe B. Laval: Telescoping Sums (PDF; 95.1 kB) - Derivation of a general theorem for telescoping sums of the form ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n\cdot (n+k)}}}$