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Analytic continuation of power series

If Σ a x converges for small complex x and can be analytically continued along some path from x = 0 to the point x = 1, then the sum of the series can be defined to be the value at x = 1. This value may depend on the choice of path. One of the first examples of potentially different sums for a divergent series, using analytic continuation, was given by Callet, who observed that if 1 ≤ m < n {\displaystyle 1\leq m<n} then

1 − x m 1 − x n = 1 + x + ⋯ + x m − 1 1 + x + … x n − 1 = 1 − x m + x n − x n + m + x 2 n − … {\displaystyle {\frac {1-x^{m}}{1-x^{n}}}={\frac {1+x+\dots +x^{m-1}}{1+x+\dots x^{n-1}}}=1-x^{m}+x^{n}-x^{n+m}+x^{2n}-\dots }

Evaluating at x = 1 {\displaystyle x=1}, one gets

1 − 1 + 1 − 1 + ⋯ = m n. {\displaystyle 1-1+1-1+\dots ={\frac {m}{n}}.}

However, the gaps in the series are key. For m = 1, n = 3 {\displaystyle m=1,n=3} for example, we actually would get

1 − 1 + 0 + 1 − 1 + 0 + 1 − 1 + ⋯ = 1 3 {\displaystyle 1-1+0+1-1+0+1-1+\dots ={\frac {1}{3}}}, so different sums correspond to different placements of the 0 {\displaystyle 0} 's.

Another example of analytic continuation is the divergent alternating series ∑ k ≥ 0 (− 1) k + 1 1 2 k − 1 (2 k k) = 1 + 2 − 2 + 4 − 10 + 28 − 84 + 264 − 858 + 2860 − 9724 + ⋯ {\displaystyle \sum _{k\geq 0}(-1)^{k+1}{\frac {1}{2k-1}}{\binom {2k}{k}}=1+2-2+4-10+28-84+264-858+2860-9724+\cdots } which is a sum over products of Γ {\displaystyle \Gamma } -functions and Pochhammer's symbols. Using theduplication formula of the Γ {\displaystyle \Gamma } -function, it reduces toa generalized hypergeometric series … = ∑ k ≥ 0 (− 4) k (− 1 / 2) k k ! = 1 F 0 (− 1 / 2;; − 4) = 5. {\displaystyle \ldots =\sum _{k\geq 0}(-4)^{k}{\frac {(-1/2)_{k}}{k!}}={}_{1}F_{0}(-1/2;;-4)={\sqrt {5}}.}