Math for Non-Geeks/Computation rules for series

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We learned that series are in some sense "infinite sums". Do the same rules as for finite sums also apply to infinite sums? Like removing braces (associative law) and re-arranging terms (commutative law)? The answer is: Generally no. But in certain cases yes! The upcoming articles will tell you when the answer is yes and when it is no. A small spoiler ahead: adding series and multiplying them by a constant is always allowed - provided that the series converges.

Overview

Computation rules

In the article "Limit theorems" we proved the sum rule for sequences $\underset{n\to \mathrm{\infty }}{\lim }(a_n+b_n)=\underset{n\to \mathrm{\infty }}{\lim }{a}_n+\underset{n\to \mathrm{\infty }}{\lim }{b}_n$ , which holds if $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ converge. These also hold for convergent series, since a series is just a sequence of partial sums. More precisely, if${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ and ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ converge and $\lambda \in ℝ$, then:

In addition, a series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges, whenever the even and odd subsequence ${\sum }_{k=1}^{\mathrm{\infty }}{a}_{2k}$ and ${\sum }_{k=1}^{\mathrm{\infty }}{a}_{2k-1}$ converge. And there is

More generally, within a convergent series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$, we can set brackets and split

Here, $(j_l{)}_{l\in \mathbb{N}}$ is the strictly monotonically increasing sequence of natural numbers with $j_1=1$ where $j_l$ indexes the first summand within a bracket-sum. Conversely, for a divergent series ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$, we also have divergence of ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$.

What goes wrong with series

For partial sums we have $\left(\sum _{k=1}^n{a}_k\right)\cdot \left(\sum _{k=1}^n{b}_k\right)\ne \sum _{k=1}^n{a}_k{b}_k$. Multiplying two series is way harder: sometimes it works and sometimes not. We will cover the details later.

There is no general associative or commutative law for series: For finite sums, one may re-arrange terms and set brackets arbitrarily and still get the same result. For infinite sums (series), this does not work in general. However, there are indicators that tell us when it works and when not.

The associative law

Why there is no associative law

For finite sums, the "Assoziativgesetzes der Addition" (German) allows to set brackets arbitrarily. For instance

Analogously

For "infinite sums", we need to pay attention: consider

The sequence of partial sums for this series is:

Which means, the partial sums "jump" between $0$ and $1$ , so the series diverges ($0$ and $1$ are accumulation points). Setting brackets can, however, lead to a series converging to 0:

So if a series diverges, we cannot simply set brackets as we wish! For convergent series, the same holds true, since we can turn the series converging to 0 above into a divergent series by removing brackets: for ${\sum }_{k=1}^{\mathrm{\infty }}((-1{)}^{2k}+(-1{)}^{2k+1})$ (which converges to 0), removing brackets yields ${\sum }_{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}$ (which diverges).

Example: where we can set brackets

Consider the converging series ${\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}$, which is an infinite sum $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$ . The corresponding sequence of partial sums is

What happens if we set brackets? We could, for instance, conclude every two neighbouring elements: $(1+\frac{1}{2})+(\frac{1}{4}+\frac{1}{8})+\dots$. This leads to the series ${\sum }_{k=0}^{\mathrm{\infty }}(2^{-2k}+2^{-(2k+1)})$. The corresponding sequence of partial sums is

This is a subsequence of the original sequence of partial sums. Now, since the series ${\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}$ converges, the sequence of its partial sums converges and hence every subsequence converges as well (and to the same limit. So ${\sum }_{k=0}^{\mathrm{\infty }}(2^{-2k}+2^{-(2k+1)})$ has the same limit as the original series. In this case, we case set brackets as we wish!

When can brackets be set and when not?

If we set brackets within a series and then consider the "bracketed series", then the partial sum sequence of the "bracketed series" is a subsequence of the original sequence of partial sums. Now

• If a sequence converges, every subsequence converges.
• If a subsequence diverges, the original sequence also diverges.

Since setting brackets leads to a subsequence of partial sums, we have that:

• Within converging series, brackets can be set arbitrarily.
• Within diverging series, brackets can be removed arbitrarily.

Or, concluded in a theorem:

What?! The sum of all natural numbers is equal to -1/12?

There are several Youtube videos and also some articles (here is a German one ) where people claim to have proven that the sum of all natural numbers equals $-\frac{1}{12}$:

This is obviously wrong! For the series above, the sequence of partial sums diverges quadratically to $\mathrm{\infty }$. It does not even attain any negative value. How do people then come up with the $-\frac{1}{12}$, then? The answer is: "by violating the associative law". All we have to do is to set brackets in divergent series (which is not allowed). Recall the sum formula for the geometric series:

In addition, for the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k$ we have the identity

If we divide this equation by $2$, we get $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k=\frac{1}{4}$. Subtracting it from the original series $\sum _{k=1}^{\mathrm{\infty }}k$ yields

After all those illegal steps, we get

q.e.d. (or rather w.t.f.)

Outlook: Series and vector spaces

For series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ and ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ as well as $\lambda \in ℝ$ we have the following computational rules:

In linear algebra, the notion of a vector space is introduced, which is roughly speaking "a set of elements, where we are allowed to add any two elements or multiply an element by a constant $\lambda \in ℝ$". The set of all real valued sequence $V=\{(a_n{)}_{n\in \mathbb{N}}\in V:a_n\in ℝ\}$ is such a set, where we can add elements or multiply by a constant. So it is a vector space. The subset $U=\{(a_n{)}_{n\in \mathbb{N}}:{\sum }_{k=1}^{\mathrm{\infty }}{a}_k\text{ converges}\}$ which includes all sequences $(a_n{)}_{n\in \mathbb{N}}$, for which the series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges is a subset of $V$ , which is a vector space again (we do not leave it by adding elements or multiplying by a constant). Such a subset is also called a subspace. The map $f:U\to ℝ:(a_n{)}_{n\in \mathbb{N}}\mapsto {\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ assigning each $(a_n{)}_{n\in \mathbb{N}}$ the limit of the series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ preserves addition and scalar multiplication: Adding two series leads to addition of the limits. Scaling the series by a constant leads to a scaling of the limit by the same constant. maps which preserve addition and scaling are also called linear maps, so y$f$ is a linear map.

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