// Twitter Cards // Prexisting Head The Biologist Is In: Mathematical Recreations : Summation of an Infinite Series

Thursday, January 16, 2014

Mathematical Recreations : Summation of an Infinite Series

My studies in biology have been aided by a solid grasp of mathematics.   I will often mathematically model some system I am working with, to try and predict how it is going to behave.   When I find something that looks mathematically implausible, I like to examine it in some detail to figure out why it looks implausible.

Today I found an article with a video explaining how the sum of all natural numbers is \( -\frac{1}{12} \).
\( \sum_{i=1}^\infty = 1+2+3+4+5+6+7+8+9 … \infty = -\frac{1}{12} \).

Sometimes my impression of implausibility is caused by my limited understanding, but this is definitely not always the case, and I find it worthwhile to explore the source of my unease at a particular statement.


The provided argument relies on summing a few simple series:
\( S_{1} = 1-1+1-1+1-1+1-1+1 … \)
\( S_{2} = 1-2+3-4+5-6+7-8+9 …  \)

…and then using them to solve the summation of the series of interest.
\( S = 1+2+3+4+5+6+7+8+9 … \)



The first series is known as Grandi's Series.   Because the series continues to infinity, with the sum alternating between 0 and 1 as you add subsequent numbers, the sequence does not converge and therefore there is no meaning to the standard sum for this series.
\( S_{1} = 1-1+1-1+1-1+1-1+1 … \)

However, the Cesàro sum is an alternate methodology where the sum is calculated as the limit of the arithmetic mean of the first \( n \) partial sums of the series, as \( n \) goes to \( \infty \).
\( x_{1} \) = {1}; \( \overline{x_{1}} = 1 \)
\( x_{2} \) = {1, 1-1} = {1, 0}; \( \overline{x_{2}} = \frac{1}{2} \)
\( x_{3} \) = {1, 1-1, 1-1+1} = {1, 0, 1}; \( \overline{x_{3}} = \frac{2}{3} \)
\( x_{4} \) = {1, 1-1, 1-1+1, 1-1+1-1}; = {1, 0, 1, 0}; \( \overline{x_{4}} = \frac{1}{2} \)
\( x_{5} \) = {1, 1-1, 1-1+1, 1-1+1-1, 1-1+1-1+1} = {1, 0, 1, 0, 1}; \( \overline{x_{5}} = \frac{3}{5} \)
\( x_{6} \) = {1, 1-1, 1-1+1, 1-1+1-1, 1-1+1-1+1, 1-1+1-1+1-1} = {1, 0, 1, 0, 1, 0}; \( \overline{x_{6}} = \frac{1}{2} \)
\( x_{7} \) = {1, 1-1, 1-1+1, 1-1+1-1, 1-1+1-1+1, 1-1+1-1+1-1, 1-1+1-1+1-1+1} = {1, 0, 1, 0, 1, 0, 1}; \( \overline{x_{7}} = \frac{4}{7} \)

…which gives us the Cesàro sequence : 1, \( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{1}{2} \), \( \frac{3}{5} \), \( \frac{1}{2} \), \( \frac{4}{7} \), \( \frac{1}{2} \), \( \frac{5}{9} \), \( \frac{1}{2} \), \( \frac{6}{11} \), \( \frac{1}{2} \), \( \frac{10}{19} \),…

… which alternates between \( \frac{1}{2} \) and a series (1, \( \frac{2}{3} \), \( \frac{3}{5} \), \( \frac{4}{7} \), \( \frac{5}{9} \), \( \frac{6}{11} \), \( \frac{7}{13} \), \( \frac{8}{15} \), \( \frac{9}{17} \), \( \frac{10}{19} \), … ) which converges to \( \frac{1}{2} \).

… so the resulting Cesàro sum is \( \frac{1}{2} \).

There is no sum to this series, only a Cesàro sum.
\( S_{Cesàro 1} = \frac{1}{2} \)



Solving the sum of the second series is a little bit more complicated:
\( S_{2} = 1-2+3-4+5-6+7-8+9 … \)

First add the series to itself:
\( 2S_{2} = S_{2}+S_{2} \)

… but offset each number in the second series by one position:
\( 2S_{2} = (1)+(-2+1)+(3-2)+(-4+3)+(5-4)+(-6+5)+(7-6)+(-8+7)+(9-8) … \)

… which simplifies to:
\( 2S_{2} = 1-1+1-1+1-1+1-1+1 … \)

So, two times our second series gives us our first series.   (Our first series did not have a sum, but we'll go ahead as if it had.)
\( 2S_{2} = S_{Cesàro 1} = \frac{1}{2} \)
$$ S_{2} = \frac{S_{Cesàro 1}}{2} = \frac{1}{4} $$



Solving the third series is then rather simple.
\( S = 1+2+3+4+5+6+7+8+9 … \)

We subtract \( S_{2} \) from \( S \) and then simplify:
\( S-S_{2} = (1+2+3+4+5+6+7+8+9 … \infty)-(1-2+3-4+5-6+7-8+9 … ) \)
\( S-S_{2} = (1-1)+(2+2)+(3-3)+(4+4)+(5-5)+(6+6)+(7-7)+(8+8)+(9-9) … ) \)
\( S-S_{2} = (0)+(4)+(0)+(8)+(0)+(12)+(0)+(16)+(0) … ) \)
\( S-S_{2} = 4+8+12+16+20+24+28+32+36 … \)
\( S-S_{2} = 4(1+2+3+4+5+6+7+8+9… ) \)
\( S-S_{2} = 4S \)
\( S-\frac{1}{4} = 4S \)
\( \frac{-1}{4} = 3S \)
\( S=\frac{-1}{12} \)
And thus, the sum of all natural numbers is equal to \( \frac{-1}{12} \).


So, what's wrong with all of that?

Some basics of series math.

The definition of a series:
\( \sum_{i=1}^\infty a_{n} = a_{1}+a_{2}+a_{3}+ … \)

A sequence of partial sums is associated with each series:
\( S_{k} = \sum_{i=1}^k a_{n} = a_{1}+a_{2}+a_{3}+ … +a_{k} \)

The series \( \sum_{i=1}^\infty a_{n} \) converges to the limit \( L \) iff (if and only if) the sequence \( S_{k} \) converges to \( L \), otherwise it is divergent.

A sum of a series for a divergent series does not exist.

For the series described in the argument:
\( S_{1} = 1-1+1-1+1-1+1-1+1 … \)
\( S_{2} = 1-2+3-4+5-6+7-8+9 …  \)
\( S = 1+2+3+4+5+6+7+8+9 … \)

All three sequences are divergent, so none of them are summable.

Some divergent series can be analyzed by an alternate method to produce a Cesàro summation, but this is distinct from the typical summation used in the described proof and only one of the series is Cesàro summable.

The Cesàro sequence for \(S_{1}\) is {\(1,\frac{1}{2},\frac{2}{3},\frac{1}{2},\frac{3}{5},\frac{1}{2},\frac{4}{7},\frac{1}{2},\frac{5}{9},\frac{1}{2},\frac{6}{11},\frac{1}{2},\frac{10}{19},…\)}. The sequence converges to \(\frac{1}{2}\), so \(S_{1}\) is Cesàro summable.

The Cesàro sequence for \(S_{2}\) is {\(1,0,\frac{2}{3},0,\frac{3}{5},0,\frac{4}{7},0,\frac{5}{9},0,\frac{6}{11},0,\frac{10}{19},…\)}. Since this sequence does not converge, \(S_{2}\) is not Cesàro summable.

The Cesàro sequence for \(S\) is {\(1,2,\frac{10}{3},5,7,\frac{28}{3},12,…\)}. Since this sequence does not converge, \(S\) is not Cesàro summable.


In summary:
\(S_{1}\) is not summable, but is Cesàro summable.
\(S_{2}\) is not summable or Cesàro summable.
\(S\) is not summable or Cesàro summable.

…which makes the argument that the sum of all natural numbers is \( \frac{-1}{12} \) somewhat hard to swallow.   The 'solution' reminds me of the methods used to prove \( 1 = 2 \), which rely on unmanaged infinities hidden away from the reader.

Dealing with series automatically means we're dealing with infinities.   The intuitions developed from working with regular numbers begin to fail when working with infinities.

Some examples :
\(\infty-\infty\not=0\)
\(\infty-\infty=\infty\)
\(\infty-5=\infty\)
\(2\infty=\infty\)
\(\infty^\infty=\infty\)

There are ways to work with infinities, and different infinities have different meanings, but care must be taken that does not appear to be taken in the 'solution' of the sum of the natural numbers that was being discussed.



In particular I distrust the solution of \( S_{2} \) because their method of multiplying the series by two can be done in many different ways to yield unexpected results.

As described above:
\( 2S_{2} = (1)+(-2+1)+(3-2)+(-4+3)+(5-4)+(-6+5)+(7-6)+(-8+7)+(9-8) … \)
\( 2S_{2} = 1-1+1-1+1-1+1-1+1 … \)
\(2S_{2} = nonexistent\)
\(2S_{Cesàro 2} = \frac{1}{2}\)
Shifting over one more:
\( 2S_{2} = (1)+(-2)+(3+1)+(-4-2)+(5+3)+(-6-4)+(7+5)+(-8-6)+(9+7) … \)
\( 2S_{2} = 1-2+4-6+8-10+12-14+16 … \)
\(2S_{2} = nonexistent\)
\(2S_{Cesàro 2} = nonexistent\)
Shifting over one more yet:
\( 2S_{2} = (1)+(-2)+(3)+(-4+1)+(5-2)+(-6+3)+(7-4)+(-8+5)+(9-6) … \)
\( 2S_{2} = 1-2+3-3+3-3+3-3+3 … \)
\(2S_{2} = nonexistent\)
\(2S_{Cesàro 2} = \frac{1}{2}\)
By multiplying each element by two:
\( 2S_{2} = 2(1)+2(-2)+2(3)+2(-4)+2(5)+2(-6)+2(7)+2(-8)+2(9) … \)
\( 2S_{2} = 2-4+6-8+10-12+14-16+18 … \)
\(2S_{2} = nonexistent\)
\(2S_{Cesàro 2} = nonexistent\)
By adding a reversed order copy of itself:
\( 2S_{2} = (1\pm\infty)+(-2\pm\infty)+(3\pm\infty)+(-4\pm\infty)+(5\pm\infty)+(-6\pm\infty)+(7\pm\infty)+(-8\pm\infty)+(9\pm\infty) … \)
\(2S_{2} = nonexistent\)
\(2S_{Cesàro 2} = nonexistent\)

If a series is not summable, then multiplying it by anything is an unreasonable thing to do.   If you choose to go ahead and do so, don't expect the results you get to be coherent.