Approximaths

‘Generic’ summation II

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Consider the series:

\displaystyle 1+0-1+1+0-1+1+0-1+\dots

using the generic summation (summation “machine”):

\displaystyle \begin{aligned} s &= \mathcal{S}(1+0-1+1+0-1+1+0-1+\dots) \\ s &= 1 + \mathcal{S}(0-1+1+0-1+1+0-1+\dots) \\ s &= 1 + \mathcal{S}(-1+1+0-1+1+0-1+\dots) \end{aligned}

Sum these 3 equations term by term:

\displaystyle \begin{aligned} 3s &= 2 + \mathcal{S}(0+0+0+0+0+0+0+\dots) \\ s &= \frac{2}{3} \end{aligned}

Now apply Euler summation to the same series:

\displaystyle 1+0-1+1+0-1+1+0-1+\dots

We multiply each n-term with x^n:

\displaystyle \begin{aligned} f(x) &:= 1x^0+0x^1-1x^2+1x^3+0x^4-1x^5+1x^6+0x^7-1x^8+\dots \\ &:= 1-x^2+x^3-x^5+x^6-x^8+\dots \\ &:= (1+x^3+x^6+x^9+\dots) - (x^2+x^5+x^8+\dots) \end{aligned}

In this last step we changed the order of summation of a Taylor series. Inside its radius of convergence this step is valid. We have:

\displaystyle \begin{aligned} f(x) &= (1+x^3+x^6+x^9+\dots) - (x^2+x^5+x^8+\dots) \\ &= (1+x^3+x^6+x^9+\dots) - x^2(1+x^3+x^6+\dots) \\ &= ((x^3)^0+(x^3)^1+(x^3)^2+(x^3)^3+\dots) - x^2((x^3)^0+(x^3)^1+(x^3)^2+(x^3)^3+\dots) \\ &= \frac{1}{1-x^3} - x^2\frac{1}{1-x^3} \\ &= \frac{1-x^2}{1-x^3} \end{aligned}

Now calculate the limit (using Hospital’s rule):

\displaystyle \lim_{x\to 1} \frac{1-x^2}{1-x^3} = \frac{2}{3}

The generic summation “machine” and the Euler summation deliver the same result.

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