User:X42bn6/Working On/Telescoping series

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This article is a rewrite of Telescoping series. Feel free to make changes or discuss them at the talk page.

In mathematics, telescoping series is an informal expression referring to a series whose sum can be found by exploiting the circumstance that nearly every term cancels with a succeeding or preceding term. The method of cancelling is also known as the method of differences or the method of differencing. When doing the method of differences, the process as a verb is known as telescoping or, loosely, differencing.

It is based largely upon the following:

$\sum^N_{n=a}\left(f(n)-f(n-1)\right)=f(N)-f(a-1)$

where f is a function that is continuous over the summation period (a to N), and is suitable for negative values of n; and a is an integer where the summation starts (usually taking the value 0 or 1). Note that the final expression will not always contain two parts. Some will contain 4 or 6, or even more. It will always be an even number because one part is cancelled at the start and at the bottom.

Telescoping is possible for the following expressions:

Polynomial identities, for example:

$\frac{1}{2}\sum^N_{n=1}\left(n(n+1)-(n-1)n\right)$

Polynomials in the denominator, which can be put into partial fractions
Trigonometric functions

Alternatively, if it will follow

$\sum^N_{n=a}\left(f(n)-f(n-1)\right)=f(N)-f(a-1)$

then telescoping is possible.

1 Examples
2 Problems with telescoping

[edit] Examples

[edit] Sums of integers raised to a power

One of the most useful ways of the method of differences is working the expressions for the sums of N integers to an integer power. This requires the power to be written in such a way that can be differenced.

For example, the sum of N consecutive integers,

$\sum^N_{n=1}n$

is an arithmetic progression with first term 1 and common difference 1. The sum can be found using:

$\sum^N_{n=1} n=\frac{N}{2}(2a+(N-1)d)$

but because a, the first term, and d, the common difference are both equal to zero, this reduces to:

$\sum^N_{n=1} n=\frac{1}{2}N(N+1).$

However, by using the identity:

2 N = N (N + 1) - (N - 1) N

we can telescope this to also deduce the formula for N consecutive integers.

$\sum^N_{n=1}n$	$=\frac{1}{2}\sum^N_{n=1}(n(n+1)-(n-1)n)$
	$=\frac{1}{2}(1\cdot 2-0\cdot 1+2\cdot 3-1\cdot 2+\cdots+(N-1)N-(N-2)(N-1)+N(N+1)-(N-1)N)$
	$=\frac{1}{2}N(N+1)$

Similarly,

$\sum^N_{n=1}n^2$

Consider:

$n^3-(n-1)^3\equiv 3n^2-3n+1$

The left-hand side can be telescoped. The right hand side cannot be telescoped, but can simply be written as sums.

For the left-hand side,

$\sum^2_{n=r}(r^3-(r-1)^3)$	$=n^3-(n-1)^3+(n-1)^3-(n-2)^3+\cdots+2^3-1^3$
	$= n 3 - 1$

Therefore,

$n 3 - 1$	$=\sum^N_{n=1}(3n^2-3n+1)$
	$=\sum^N_{n=1}3n^2-\sum^N_{n=1}3n+\sum^N_{n=1}1$
	$=3\sum^N_{n=1}n^2-3(\frac{1}{2}n(n+1))+n$
$3\sum^N_{n=1}n^2$	$=n^3-1-3(\frac{1}{2}n(n+1))-n$
	$=n^3-n-\frac{3n}{2}(n+1)-1$
	$=n(n^2-1-\frac{3}{2}(n+1))-1$

[edit] Polynomials in the denominator

Consider the expression:

$\sum^N_{n=0} \frac{1}{(n+1)(n+2)}.$

By partial fractions, we get:

$\sum^N_{n=0}\left(\frac{1}{n+1}-\frac{1}{n+2}\right).$

Then, by writing out the first few terms and the last few terms, we get:

$\sum^N_{n=0}\frac{1}{(n+1)(n+2)}$	$=\sum^N_{n=0}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
	$=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{N-1}-\frac{1}{N}+\frac{1}{N}-\frac{1}{N+1}+\frac{1}{N+1}-\frac{1}{N+2}$
	$=\frac{1}{1}-\frac{1}{N+2}$
	$=1-\frac{1}{N+2}$

[edit] Trigonometric functions

Many trigonometric functions also admit representation as a difference, which allows telescoping between the consequent terms.

$\sum_{n=1}^N \sin\left(n\right)$	$= \sum_{n=1}^N \frac{1}{2} \csc\left(\frac{1}{2}\right) \left(2\sin\left(\frac{1}{2}\right)\sin\left(n\right)\right)$
	$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \sum_{n=1}^N \left(\cos\left(\frac{2n-1}{2}\right)-\cos\left(\frac{2n+1}{2}\right)\right)$
	$=\frac{1}{2} \csc\left(\frac{1}{2}\right) \left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{2N+1}{2}\right)\right).$

[edit] Problems with telescoping

However, notice how the expression:

$\sum^N_{n=0}\frac{2n+3}{(n+1)(n+2)}$

will work by differencing. By splitting into partial fractions and telescoping, we get:

$\sum^N_{n=0}\frac{2n+3}{(n+1)(n+2)}$	$=\sum^N_{n=0}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)$
	$=\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{N-1}+\frac{1}{N}+\frac{1}{N}+\frac{1}{N+1}+\frac{1}{N+1}+\frac{1}{N+2}$
	$=\infty$

Which is untrue, as the sum of N numbers can be found manually, and is finite. The reason is that the plus sign in the summation expression causes telescoping to fail.

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User:X42bn6/Working On/Telescoping series

From Wikipedia, the free encyclopedia

Contents

[edit] Examples

[edit] Sums of integers raised to a power

[edit] Polynomials in the denominator

[edit] Trigonometric functions

[edit] Problems with telescoping

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