Alternating series (Euler)

Euler's alternating series is a mathematical paradox . You deal with divergent series that appear to be convergent. Furthermore, Leonhard Euler established a relationship between the alternating series of exponentiated natural numbers and those of the exponentiated reciprocals of natural numbers . One explanation of the paradox lies in the rearrangement of rows .

The series of alternating natural numbers

When it's your turn

${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}(k+1)=1-2+3-4+\ldots }$

with the partial sums

${\displaystyle s_{n}=\sum _{k=0}^{n}(-1)^{k}(k+1)=1-2+3-4+\ldots +(-1)^{n}\cdot (n+1)}$

If you look at it, you can see that this diverges , since the partial sums form a sequence of the whole numbers (1, -1, 2, -2, 3, -3, ...) sorted according to the amount . In addition , a series only converges when at least the sequence of the summands is a zero sequence .

The following rearrangement is therefore not legitimate, since just rearranging absolutely convergent series has no effect on the sum of the series. In some cases, a simple convergence is sufficient.

But since modern analysis, and with it the concept of convergence, was first practiced by Leonhard Euler and Augustin Louis Cauchy , this derivation is a reflection of what was then considered inexplicably paradoxical.

Be now . Then ${\displaystyle s=\lim _{n\to \infty }s_{n}}$

${\displaystyle {\begin{array}{rclllll}4s&=&&(1-2+3-4+\cdots )&+(1-2+3-4+\cdots )&+(1-2+3-4+\cdots )&+(1-2+3-4+\cdots )\\&=&&(1-2+3-4+\cdots )&+1+(-2+3-4+5-\cdots )&+1+(-2+3-4+5-\cdots )&-1+(3-4+5-6+\cdots )\\&=&1+[&(1-2-2+3)&+(-2+3+3-4)&+(3-4-4+5)&+(-4+5+5-6)+\cdots ]\\&=&1+[&0+0+0+0+\cdots ]\\4s&=&1\end{array}}}$

and therefore applies

${\displaystyle 1-2+3-4+\cdots ={\frac {1}{4}}.}$

Cauchy product

An equally paradoxical equation is generated by the Grandi series

e _n = Σ (−1) ⁿ = 1 - 1 + 1 - 1 +… ( 1, 0, 1, 0, 1, 0, ... ),

for which applies to a similarly elegant rearrangement . If you move away from the usual definition of a sum and ask yourself the question "What should the result of this be?", You get two possible results: ${\displaystyle \lim _{n\to \infty }e_{n}={\frac {1}{2}}}$

(1 - 1) + (1 - 1) + (1 - 1) +… = 0 + 0 + 0 +… = 0 and

1 + (−1 + 1) + (−1 + 1) + (−1 + 1) +… = 1 + 0 + 0 + 0 +… = 1.

Of course, according to today's understanding, it is to be taken ad absurdum if one shows that

S = 1 - 1 + 1 - 1 +…, so

1 - S = 1 - (1 - 1 + 1 - 1 +…) = 1 - 1 + 1 - 1 +… = S is.

The Cauchy product of the Grandi series with itself, however, surprisingly creates the explicitly represented follow-up link

${\displaystyle {\begin{array}{rcl}c_{n}&=&\displaystyle \sum _{k=0}^{n}a_{k}b_{n-k}=\sum _{k=0}^{n}(-1)^{k}(-1)^{n-k}\\[1em]&=&\displaystyle \sum _{k=0}^{n}(-1)^{n}=(-1)^{n}(n+1)\end{array}}.}$

The series over c _n is then

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}(n+1)=1-2+3-4+\cdots .}$

Euler's power series

In remarks on a beautiful relationship between real and reciprocal power series , Leonhard Euler devoted all his attention to the two series

${\displaystyle \sum _{k=0}^{\infty }(k+1)^{m}(-1)^{k}=1-2^{m}+3^{m}-4^{m}+\cdots }$ (1)

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(k+1)^{n}}}(-1)^{k}=1-{\frac {1}{2^{n}}}+{\frac {1}{3^{n}}}-{\frac {1}{4^{n}}}+\cdots }$ (2),

which are to be chosen arbitrarily. ${\displaystyle m,n\in \mathbb {N} }$

The real power series

In his remarks Euler tries not to regard the series as sums, but rather to equate them with an analytically identical expression. They help in deriving higher potencies. It only became clear later that the expressions are actually only partially identical.

It begins with the relation (3), which can be easily obtained via a Taylor expansion around x _o = 0 or by division in writing. For x = 1, we get the Grandi series with its paradoxical result. ${\displaystyle P_{0}(x)={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+\cdots =\sum _{k=0}^{\infty }(-1)^{k}\cdot x^{k}}$

He also cites the following recursive formation rule to determine the higher potencies

${\displaystyle P_{n+1}(x)={\frac {\mathrm {d} }{\mathrm {d} x}}\left(x\cdot P_{n}(x)\right)}$from what

${\displaystyle {\begin{aligned}P_{1}(x)&=1-2\cdot x+3\cdot x^{2}-4\cdot x^{3}+\cdots &=&{\frac {1}{(1+x)^{2}}}\\P_{2}(x)&=1-2^{2}\cdot x+3^{2}\cdot x^{2}-4^{2}\cdot x^{3}+\cdots &=&{\frac {1-x}{(1+x)^{3}}}\\P_{3}(x)&=1-2^{3}\cdot x+3^{3}\cdot x^{2}-4^{3}\cdot x^{3}+\cdots &=&{\frac {1-4x+x^{2}}{(1+x)^{4}}}\\P_{4}(x)&=1-2^{4}\cdot x+3^{4}\cdot x^{2}-4^{4}\cdot x^{3}+\cdots &=&{\frac {1-11x+11x^{2}-x^{3}}{(1+x)^{5}}}\\P_{5}(x)&=1-2^{5}\cdot x+3^{5}\cdot x^{2}-4^{5}\cdot x^{3}+\cdots &=&{\frac {1-26x+66x^{2}-26x^{3}+x^{4}}{(1+x)^{6}}}\\P_{6}(x)&=1-2^{6}\cdot x+3^{6}\cdot x^{2}-4^{6}\cdot x^{3}+\cdots &=&{\frac {1-57x+302x^{2}-302x^{3}+57x^{4}-x^{5}}{(1+x)^{7}}}\\\vdots \\P_{m}(x)&=1-2^{m}\cdot x+3^{m}\cdot x^{2}-4^{m}\cdot x^{3}+\cdots \\\end{aligned}}}$
open up.

For P ₁ (1) there is accordingly the alternating series of integers given above and for P _m (1) the series (1).

Attempts to explain

As already mentioned above, rearrangement is permissible at least for suitable convergent series, but at most for absolutely convergent series.

However, the main reason lies in equation (3). Because equality would only be given for, but it can never be achieved because infinity is unattainable. ${\displaystyle n=\infty }$

In addition, with a development from at some point you will be forced to terminate, so that there is always a residual term that disrupts the equality. ${\displaystyle \textstyle {\frac {1}{1+x}}}$

${\displaystyle {\begin{aligned}{\frac {1}{1+x}}&=\left(\sum _{k=0}^{n}(-1)^{k}\cdot x^{k}\right)+(-1)^{n+1}\cdot {\frac {x^{n+1}}{1+x}}={\begin{cases}1-{\frac {1}{2}},&{\text{wenn }}n{\text{ gerade, }}x=1\\0+{\frac {1}{2}},&{\text{wenn }}n{\text{ ungerade, }}x=1\\\end{cases}}\\\\&\neq \sum _{k=0}^{n}(-1)^{k}\cdot x^{k}\end{aligned}}}$

This problem can only be avoided if the series is only considered for, since then for correspondingly small amounts of x or for correspondingly large n the remainder term tends towards zero. ${\displaystyle \textstyle |x|<1}$

Therefore it is only possible to consider the limit value . ${\displaystyle \textstyle \lim _{x\to 1}P_{m}(x)}$

literature

• Leonhard Euler: Translation with notes of Euler's paper: Remarks on a beautiful relation between direct as well as reciprocal power series . Ed .: The Euler Archive. E352, 2006 (English, math.dartmouth.edu [PDF; 337 kB ; accessed on 28 September 2016] French: Remarques sur un beau rapport entre les series des puissances tant directes que reciproques . Translated by Lucas Willis, Thomas J Osler, - first in: Memoires de l'academie des sciences de Berlin , 17, pp. 83-106).

Individual evidence

1. ↑ jstor.org
2. ^ Leonhard Euler: Translation with notes of Euler's paper: Remarks on a beautiful relation between direct as well as reciprocal power series . Ed .: The Euler Archive. E352, 2006 (English, math.dartmouth.edu [PDF; 337 kB ; accessed on 28 September 2016] French: Remarques sur un beau rapport entre les series des puissances tant directes que reciproques . Translated by Lucas Willis, Thomas J Osler, Memoires de l'academie des sciences de Berlin , 17, pp. 83-106).