Computation rules for series – "Math for Non-Geeks"

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:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}(a_k+b_k)& =\sum _{k=1}^{\mathrm{\infty }}{a}_k+\sum _{k=1}^{\mathrm{\infty }}{b}_k\\ \sum _{k=1}^{\mathrm{\infty }}\lambda a_k& =\lambda \sum _{k=1}^{\mathrm{\infty }}{a}_k\end{array}$

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

$\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}+\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$

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

$\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{l=1}^{\mathrm{\infty }}\left(\sum _{k=j_l}^{j_{l+1}-1}{a}_k\right)$

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.

Question: Find an example for two series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ and $\sum _{k=1}^{\mathrm{\infty }}{b}_k$, where $\left(\sum _{k=1}^{\mathrm{\infty }}{a}_k\right)\cdot \left(\sum _{k=1}^{\mathrm{\infty }}{b}_k\right)\ne \sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$

An example is $a_1=2,a_2=1,a_k=0$ for $k\ge 3$ and $b_1=1,b_k=0$ for $k\ge 2$: $\begin{array}{cc}\left(\sum _{k=1}^{\mathrm{\infty }}{a}_k\right)\cdot \left(\sum _{k=1}^{\mathrm{\infty }}{b}_k\right)& =(2+1+0+\dots )\cdot (1+0+\dots )=3\cdot 1=3\\ \sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k& =2\cdot 1+1\cdot 0+0\cdot 0+\dots =2\end{array}$

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 sum rule

Proof of the sum rule

Theorem (sum rule for series)

Let ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ and ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ be two convergent series. Then

$\sum _{k=1}^{\mathrm{\infty }}(a_k+b_k)=\sum _{k=1}^{\mathrm{\infty }}{a}_k+\sum _{k=1}^{\mathrm{\infty }}{b}_k$

Proof (sum rule for series)

There is:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}(a_k+b_k)& =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n(a_k+b_k)\\ & =\underset{n\to \mathrm{\infty }}{lim}(\sum _{k=1}^n{a}_k+\sum _{k=1}^n{b}_k)\\ & \downarrow \underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=\underset{n\to \mathrm{\infty }}{lim}{a}_n+\underset{n\to \mathrm{\infty }}{lim}{b}_n\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n{a}_k+\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n{b}_k\\ & =\sum _{k=1}^{\mathrm{\infty }}{a}_k+\sum _{k=1}^{\mathrm{\infty }}{b}_k\end{array}$

We are allowed to use the limit theorem $\underset{n\to \mathrm{\infty }}{lim}({\sum }_{k=1}^n{a}_k+{\sum }_{k=1}^n{b}_k)=\underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{a}_k+\underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{b}_k$ since the series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ and ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ converge, so the sequences of partial sums $\underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{a}_k$ and $\underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{b}_k$ converge, i.e. their limits exist.

Example problems: sum rule for series

Exercise (sum rule for series)

Compute the value of ${\sum }_{k=0}^{\mathrm{\infty }}\frac{1+(-1{)}^k}{2^k}$.

Solution (sum rule for series)

There is

$\begin{array}{cc}\sum _{k=0}^{\mathrm{\infty }}\frac{1+(-1{)}^k}{2^k}& =\sum _{k=0}^{\mathrm{\infty }}[\frac{1}{2^k}+\frac{(-1{)}^k}{2^k}]\\ & =\sum _{k=0}^{\mathrm{\infty }}[{\left(\frac{1}{2}\right)}^k+{(-\frac{1}{2})}^k]\\ & \downarrow \sum _{k=0}^{\mathrm{\infty }}(a_k+b_k)=\sum _{k=0}^{\mathrm{\infty }}{a}_k+\sum _{k=0}^{\mathrm{\infty }}{b}_k\\ & =\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^k+\sum _{k=0}^{\mathrm{\infty }}{(-\frac{1}{2})}^k\\ & \downarrow \begin{array}{cc}\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^k& =\frac{1}{1-\frac{1}{2}}=2\\ \sum _{k=0}^{\mathrm{\infty }}{(-\frac{1}{2})}^k& =\frac{1}{1-(-\frac{1}{2})}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\end{array}\\ & =2+\frac{2}{3}=\frac{8}{3}\end{array}$

We are allowed to use the sum rule, since the partial series converge.

The factor rule

Proof of the factor rule

Theorem (factor rule for series)

Let ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ be a convergent series and $\lambda \in ℝ$ a real number. Then

$\sum _{k=1}^{\mathrm{\infty }}\lambda a_k=\lambda \cdot \sum _{k=1}^{\mathrm{\infty }}{a}_k$

Proof (factor rule for series)

There is:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}\lambda a_k& =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\lambda a_k\\ & =\underset{n\to \mathrm{\infty }}{lim}(\lambda \cdot \sum _{k=1}^n{a}_k)\\ & \downarrow \underset{n\to \mathrm{\infty }}{lim}\lambda \cdot a_n=\lambda \cdot \underset{n\to \mathrm{\infty }}{lim}{a}_n\\ & =\lambda \cdot \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n{a}_k\\ & =\lambda \cdot \sum _{k=1}^{\mathrm{\infty }}{a}_k\end{array}$

We are allowed to use $\underset{n\to \mathrm{\infty }}{lim}(\lambda \cdot {\sum }_{k=1}^n{a}_k)=\lambda \cdot \underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{a}_k$ since ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges, so the limit $\underset{n\to \mathrm{\infty }}{lim}{\sum }_{k=1}^n{a}_k$ exists.

Example problems: factor rule for series

Exercise

Compute ${\sum }_{k=4}^{\mathrm{\infty }}\frac{1}{2^k}$.

Solution

There is

$\begin{array}{cc}\sum _{k=4}^{\mathrm{\infty }}\frac{1}{2^k}& =\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2^{k+4}}\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2^4}\cdot \frac{1}{2^k}\\ & \downarrow \sum _{k=0}^{\mathrm{\infty }}\lambda a_k=\lambda \cdot \sum _{k=0}^{\mathrm{\infty }}{a}_k\\ & =\frac{1}{2^4}\cdot \sum _{k=0}^{\mathrm{\infty }}\frac{1}{2^k}\\ & \downarrow \sum _{k=0}^{\mathrm{\infty }}\frac{1}{2^k}=\frac{1}{1-\frac{1}{2}}=2\\ & =\frac{1}{2^4}\cdot 2=\frac{1}{2^3}=\frac{1}{8}\end{array}$

The series converges, so we are allowed to use the factor rule.

The splitting rule

Proof of the splitting rule

Theorem (splitting rule for series)

Let $(a_k{)}_{k\in \mathbb{N}}$ be a sequence. If $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$ converge, then $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges as well, and there is:

$\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}+\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$

Proof (splitting rule for series)

This is a consequence of the sum rule above. We take a look at the series $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$ . They are given by the partial sum series

$\begin{array}{cc}{\left(\sum _{k=1}^n{a}_{2k-1}\right)}_{n\in \mathbb{N}}& =(a_1,a_1+a_3,a_1+a_3+a_5,\dots )\\ {\left(\sum _{k=1}^n{a}_{2k}\right)}_{n\in \mathbb{N}}& =(a_2,a_2+a_4,a_2+a_4+a_6,\dots )\end{array}$

We can create two new sequences $(b_k{)}_{k\in \mathbb{N}}$ and $(c_k{)}_{k\in \mathbb{N}}$, by extracting the elements from $(a_{2k-1}{)}_{k\in \mathbb{N}}$ and $(a_{2k}{)}_{k\in \mathbb{N}}$ and "filling up the gaps" with zeros

$\begin{array}{cc}(b_k{)}_{k\in \mathbb{N}}& =(a_1,0,a_3,0,a_5,\dots )\\ (c_k{)}_{k\in \mathbb{N}}& =(0,a_2,0,a_4,0,\dots )\end{array}$

The corresponding sequences of partial sums are then

$\begin{array}{cc}{\left(\sum _{k=1}^n{b}_k\right)}_{n\in \mathbb{N}}& =(a_1,a_1+0,a_1+0+a_3,a_1+0+a_3+0,\dots )\\ {\left(\sum _{k=1}^n{c}_k\right)}_{n\in \mathbb{N}}& =(0,0+a_2,0+a_2+0,0+a_2+0+a_4,\dots )\end{array}$

Since $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$ converge the series $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ and $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ converge as well, with

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}{b}_k& =\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}\\ \sum _{k=1}^{\mathrm{\infty }}{c}_k& =\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}\end{array}$

The sum rule implies convergence of $\sum _{k=1}^{\mathrm{\infty }}(b_k+c_k)$ . Now $a_k=b_k+c_k$ for all $k\in \mathbb{N}$. Hence, $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ has to converge as well, where

$\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{k=1}^{\mathrm{\infty }}(b_k+c_k)=\sum _{k=1}^{\mathrm{\infty }}{b}_k+\sum _{k=1}^{\mathrm{\infty }}{c}_k=\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}+\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$

Question: Does the converse also hold true? Meaning, if $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges, then also $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}$ and $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}$ converge?

No: The alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k}$ converges. But its even partial series $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}=\sum _{k=1}^{\mathrm{\infty }}-\frac{1}{2k}$ diverges, as it is half of the diverging harmonic series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ .- The subseries of odd elements $\sum _{k=1}^{\mathrm{\infty }}{a}_{2k-1}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k-1}$ diverges, as well.

Example problems splitting rule

Exercise (splitting rule for series)

Let $a_n=\{\begin{array}{cc}(\frac{1}{3}{)}^n& \text{ for }n=2k,\\ 0& \text{ for }n=2k-1.\end{array}$ . Compute the value of $\sum _{n=0}^{\mathrm{\infty }}{a}_n$.

Solution (splitting rule for series)

There is

$\begin{array}{cc}\sum _{n=0}^{\mathrm{\infty }}{a}_n& =\sum _{k=0}^{\mathrm{\infty }}[a_{2k}+a_{2k+1}]\\ & \downarrow \sum _{k=0}^{\mathrm{\infty }}{a}_k=\sum _{k=0}^{\mathrm{\infty }}{a}_{2k}+\sum _{k=0}^{\mathrm{\infty }}{a}_{2k-1}\\ & =\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^{2k}+\sum _{k=0}^{\mathrm{\infty }}0\\ & =\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{9}\right)}^k+0\\ & \downarrow \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{9}\right)}^k=\frac{1}{1-\frac{1}{9}}=\frac{9}{8}\\ & =\frac{9}{8}\end{array}$

As the series converges, we can use the computational rules.

The associative law

Why there is no associative law

For finite sums, the associative law allows putting brackets arbitrarily. For instance

$\begin{array}{cc}1-1+1-1+1-1& =(((((1-1)+1)-1)+1)-1)\\ & =((((0+1)-1)+1)-1)\\ & =\dots =0\end{array}$

Analogously

$(1-1)+(1-1)+(1-1)=0+0+0=0$

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

$\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}=1-1+1-1\pm \dots$

The sequence of partial sums for this series is:

$\begin{array}{cc}{(\sum _{k=1}^n(-1{)}^{k+1})}_n& =(1,1-1,1-1+1,1-1+1-1,\dots )\\ & =(1,0,1,0,\dots )\end{array}$

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:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}((-1{)}^{2k}+(-1{)}^{2k+1})& =(1-1)+(1-1)+(1-1)+(1-1)+\dots \\ & =0+0+0+0+\dots =0\end{array}$

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).

Question: Can one also set brackets in a way that $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}$ converges to $1$ or $-1$ ?

To obtain the limit $1$ , we use

$\begin{array}{cc}1+\sum _{k=1}^{\mathrm{\infty }}((-1{)}^{2k+1}+(-1{)}^{2k+2})& =1+(-1+1)+(-1+1)+(-1+1)+\dots \\ & =1+0+0+0+\dots =1\end{array}$

Achieving $-1$ does not work, since the partial sums for every setting of brackets can either take the value 0 or 1.

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

${\left(\sum _{k=0}^n{2}^{-k}\right)}_{n\in \mathbb{N}}=(1,1+\frac{1}{2},1+\frac{1}{2}+\frac{1}{4},\dots )$

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

${\left(\sum _{k=0}^n(2^{-2k}+2^{-(2k+1)})\right)}_{n\in \mathbb{N}}=(1+\frac{1}{2},1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8},\dots )$

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:

Theorem (brackets in series)

If a series converges to some limit, then every series obtained from it by setting more brackets converges to the same limit. If a series diverges, then every series obtained from it by removing some brackets diverges.

Let ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ be a convergent series and $(j_l{)}_{l\in \mathbb{N}}$ a strictly monotonically increasing sequence of natural numbers with $j_1=1$. Here, $j_l$ is the index of the first summand of the $l$-th bracket. Now,

• If ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges, then ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$ converges to the same limit.
• If ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$ diverges, then ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ diverges, as well.

Proof (brackets in series)

Let ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ be a converging series. Introducing new brackets, we obtain ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$, where $(j_l{)}_{l\in \mathbb{N}}$ is a strictly monotonically increasing sequence of natural numbers with $j_1=1$. The number $j_l$is the index of the first summand of the $l$-th bracket. The corresponding sequence of partial sums now reads:

$\begin{array}{cc}{\left(\sum _{l=1}^n\left(\sum _{k=j_l}^{j_{l+1}-1}{a}_k\right)\right)}_{n\in \mathbb{N}}& =(\sum _{k=1}^{j_2-1}{a}_k,\sum _{k=1}^{j_2-1}{a}_k+\sum _{k=j_2}^{j_3-1}{a}_k,\dots )\\ & =(\sum _{k=1}^{j_2-1}{a}_k,\sum _{k=1}^{j_3-1}{a}_k,\sum _{k=1}^{j_4-1}{a}_k,\dots )\\ & ={\left(\sum _{k=1}^{j_{n+1}-1}{a}_k\right)}_{n\in \mathbb{N}}\end{array}$

This is a subsequence of the original sequence of partial sums. A subsequence converges to the same limit as the original sequence. So

• If ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges, then also the subsequence ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$ has to converge to the same limit.
• If ${\sum }_{l=1}^{\mathrm{\infty }}\left({\sum }_{k=j_l}^{j_{l+1}-1}{a}_k\right)$ diverges, then ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ cannot be convergent to any limit in the first place. So it must diverge, as well..

So in converging series, we can set and in diverging series, we can remove brackets as we wish.

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 $-\frac{1}{12}$:

$\sum _{k=1}^{\mathrm{\infty }}k=1+2+3+4+5+6+7+8+\dots =-\frac{1}{12}$

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:

$1-1+1-1+1-1+1-1\pm \dots =\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k=\frac{1}{1-(-1)}=\frac{1}{2}$

Question: What is wrong in the line above?

The sum rule for the geometric series $\sum _{k=1}^{\mathrm{\infty }}{q}^k$ only holds in case of convergence. For $q=-1$, it diverges. So the limit is not $\frac{1}{1-q}$ for $q=-1$.

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

$\begin{array}{cc}2\cdot \sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k& =2\cdot (1-2+3-4+5-6+7-8\pm \dots )\\ & =2-4+6-8+10-12+14-16\pm \dots \\ & =1+1-2-2+3+3-4-4+5+5-6-6+7+7-8-8+\dots \\ & =1+(1-2)+(-2+3)+(3-4)+(-4+5)+(5-6)+(-6+7)+(7-8)+\dots \\ & =1-1+1-1+1-1+1-1+\dots \\ & =\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k=\frac{1}{2}\end{array}$

Question: And what is wrong, here?

Multiplying out and factoring out is not allowed for diverging infinite sums:

$\begin{array}{cc}2\cdot \sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k& =2\cdot (1-2+3-4+5-6+7-8\pm \dots )\\ & =2-4+6-8+10-12+14-16\pm \dots \\ & =1+1-2-2+3+3-4-4+5+5-6-6+7+7-8-8+\dots \\ & =1+(1-2)+(-2+3)+(3-4)+(-4+5)+(5-6)+(-6+7)+(7-8)+\dots \\ & =1-1+1-1+1-1+1-1+\dots \\ & =\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k=\frac{1}{2}\end{array}$

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

$\begin{array}{cc}\left(\sum _{k=1}^{\mathrm{\infty }}k\right)-\frac{1}{4}& =\sum _{k=1}^{\mathrm{\infty }}k-\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k\\ & =(1+2+3+4+5+6+7+8+\dots )-(1-2+3-4+5-6+7-8\pm \dots )\\ & =0+4+0+8+0+12+0+16+\dots \\ & =4\cdot (1+2+3+4+5+6+7+8+\dots )\\ & =4\cdot \sum _{k=1}^{\mathrm{\infty }}k\end{array}$

Question: And what is wrong with this step?

$\begin{array}{cc}\left(\sum _{k=1}^{\mathrm{\infty }}k\right)-\frac{1}{4}& =\sum _{k=1}^{\mathrm{\infty }}k-\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}k\\ & =(1+2+3+4+5+6+7+8+\dots )-(1-2+3-4+5-6+7-8\pm \dots )\\ & =0+4+0+8+0+12+0+16+\dots \\ & =4\cdot (1+2+3+4+5+6+7+8+\dots )\\ & =4\cdot \sum _{k=1}^{\mathrm{\infty }}k\end{array}$

When the series $\sum _{k=1}^{\mathrm{\infty }}k$ and $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k}k$ were subtracted element-wise, we assumed that the sum rule would hold. But this is not true for divergent series. The factor rule does also not hold true in this case, so pulling out a $4$ in $4+8+12+16+\dots =\sum _{k=1}^{\mathrm{\infty }}4k$ is not allowed.

After all those illegal steps, we get

$\begin{array}{cc}& \left(\sum _{k=1}^{\mathrm{\infty }}k\right)-\frac{1}{4}=4\cdot \sum _{k=1}^{\mathrm{\infty }}k\\ ⟺& -3\left(\sum _{k=1}^{\mathrm{\infty }}k\right)-\frac{1}{4}=0\\ ⟺& -3\left(\sum _{k=1}^{\mathrm{\infty }}k\right)=\frac{1}{4}\\ ⟺& \sum _{k=1}^{\mathrm{\infty }}k=-\frac{1}{12}\end{array}$

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:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}(a_k+b_k)& =\sum _{k=1}^{\mathrm{\infty }}{a}_k+\sum _{k=1}^{\mathrm{\infty }}{b}_k\\ \sum _{k=1}^{\mathrm{\infty }}\lambda a_k& =\lambda \sum _{k=1}^{\mathrm{\infty }}{a}_k\end{array}$

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.