Approximaths

‘Generic’ summation I

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Let imagine that we have engineered a summation ‘machine’ called \mathcal{S}() and consider the geometric series:

\displaystyle 1+x+x^2+x^3+x^4+\dots

Apply our summation ‘machine’ on this series and call the result s:

\displaystyle s = \mathcal{S}(1+x+x^2+x^3+x^4+\dots)

We assign the following two properties for the ‘machine’ \mathcal{S}:

\displaystyle \mathcal{S}(a_0+a_1+a_2+\dots) = a_0 + \mathcal{S}(a_1+a_2+\dots) \quad \text{(first property)}
\displaystyle \mathcal{S}(\sum(\alpha a_n) + \sum(\beta b_n)) = \alpha \mathcal{S}(\sum a_n) + \beta \mathcal{S}(\sum b_n)) \quad \text{(second property)}

where \alpha and \beta are constants. Equipped with the machine \mathcal{S}() and its two properties consider again the geometric series:

\displaystyle \begin{aligned} s &= \mathcal{S}(1+x+x^2+x^3+x^4+\dots) &\quad\text{(definition)}\\ s &= 1+ \mathcal{S}(x+x^2+x^3+x^4+\dots) &\quad\text{(first property)}\\ s &= 1+ x\mathcal{S}(1+x+x^2+x^3+\dots) &\quad\text{(second property)}\\ s &= 1+ xs &\quad\text{(definition)}\\ s &= \frac{1}{1-x} \end{aligned}

Consider the series:

\displaystyle 1-1+1-1+1-1+\dots

If we use traditional, ‘rigorous’, summation techniques we will conclude that this alternating series is not converging. Now apply the summation “machine” as described above:

\displaystyle \begin{aligned} s &= \mathcal{S}(1-1+1-1+1-1+\dots ) \\ s &= 1+ \mathcal{S}(-1+1-1+1-1+\dots ) \\ s &= 1 -\mathcal{S}(1-1+1-1+1-\dots ) \\ s &= 1-s \\ s &= \frac{1}{2} \end{aligned}

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