To calculate an infinite sum exactly, try antidifferencing

a repository of mathematical know-how

To calculate an infinite sum exactly, try antidifferencing

Title: *

Area of mathematics: *

A comma-separated list of areas of mathematics to which this article applies. Use ">" to tag in a subcategory. Example: Analysis > Harmonic analysis, Combinatorics

Keywords:

A comma-separated list of keywords associated with this article. Example: free group

Used in:

A comma-separated list of examples of where this technique is used. Example: Cauchy-Schwarz inequality

Parent articles:		Order

Body:

[QUICK DESCRIPTION]
If you are trying to calculate the sum $\sum_{n=1}^\infty a_n$, then sometimes it is possible to spot another sequence $(b_n)$ such that $a_n=b_n-b_{n+1}$ and $b_n\rightarrow 0$. If that is the case, then your sum is equal to $b_1$.

[PREREQUISITES]
The definition of an infinite sum.

[note contributions wanted] Probably there is more to be said on this theme [/note]

[EXAMPLE]
An infinite sum that is well known to be straightforward to calculate exactly is the sum 
[maths] \frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\frac 1{3\cdot 4}+\dots[/maths]
The usual technique for summing this is to observe that $\frac 1{n(n+1)}=\frac 1n-\frac 1{n+1}$, from which it follows that 
[maths] \frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\dots+\frac 1{N(N+1)}=1-\frac 1{N+1},[/maths]
which tends to 1 as $N$ tends to infinity.

This points to a circumstance in which an infinite sum can be evaluated exactly: we can work out the discrete analogue of an "antiderivative": that is, we have a sequence $(a_n)$ and can spot a nice sequence $(b_n)$ such that $b_n-b_{n+1}=a_n$. Of course, $b_n$ must in that case be the partial sum $a_1+\dots+a_n$ (up to an additive constant). So is this really making the more or less tautologous observation that if you have a nice formula for the partial sums and can see easily what they converge to then you are done?

It isn't quite, because we could have spotted that $\frac 1{n(n+1)}=\frac 1n-\frac 1{n+1}$ without working out any partial sums.

[GENERAL DISCUSSION]
Let us try to understand better why the above trick is not a universal method for calculating all infinite sums, by contrasting the above example with the example of the sum 
[maths] \frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+\dots [/maths]
Can we find some sequence $(b_n)$ such that $b_n-b_{n+1}=1/n^2$? It seems difficult just to spot such a sequence, so instead let us try to be systematic about it. It's not quite clear how to pick $b_1$, so let us set it to be $\sigma$. Then $\sigma-b_2=1$, so $b_2=\sigma-1$. Next, $b_2-b_3=1/4$, so $b_3=\sigma-1-1/4$. In general, we find that $b_n=\sigma-1-1/4-\dots-1/(n-1)^2$. Since we also want $b_n$ to tend to zero, this tells us what $\sigma$ must be: $\sum_{n=1}^\infty 1/n^2$, precisely the sum we were trying to calculate!

The reason the technique worked in the example above was that the partial sums turned out to have a nice formula.

[remark] A simple method of generating undergraduate mathematics exercises is to reverse the telescoping-sums idea in order to create infinite sums that can be evaluated exactly. For example, take the sequence $b_n=1/(1+n^2)$. This tends to $0$ as $n$ tends to infinity. Now let 
[maths]a_n=b_n-b_{n+1}=\frac 1{1+n^2}-\frac 1{1+(n+1)^2}=\frac{(n+1)^2-n^2}{(1+n^2)(1+(n+1)^2}=\frac{2n+1}{(n^2+1)(n^2+2n+2)}.[/maths]
The question you then ask is to calculate the sum $\sum_{n=1}^\infty\frac{2n+1}{(n^2+1)(n^2+2n+2)}$. A smart student will use partial fractions to discover that this is a telescoping sum and will end up with the answer $b_1=1/2$.[/remark]

If this method works, it is a bit like managing to calculate an integral by antidifferentiating the integrand.

This is a stub

A stub is an article that is not sufficiently complete to be interesting.

Notifications

File attachments

Changes made to the attachments are not permanent until you save this post. The first "listed" file will be included in RSS feeds.

Attach new file:

Images are larger than 640x480 will be resized. The maximum upload size is 1 MB. Only files with the following extensions may be uploaded: jpg jpeg gif png svg.

Revision information

Log message:

An explanation of the additions or updates being made to help other authors understand your motivations.

To calculate an infinite sum exactly, try antidifferencing

Recent articles

Active forum topics

Recent comments