I want to speed up the convergence of a series involving rational expressions the expression is ∑x=1∞(−1)x−x2−2x+1x4+2x2+1

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I want to speed up the convergence of a series involving rational expressions the expression is  ∑x=1∞(−1)x−x2−2x+1x4+2x2+1

euromillionsna · Accepted Answer

Here is the first step toward making the series evaluate quicker. You may need just as many terms but it will be faster. Break the series up through partial fractions. −x2−2x+1x4+2x2+1=−2(x−1)(x2+1)2−1x2+1 I will focus on the second term in the sum. ∑x=0∞(−1)xx2+1=∑n=0∞14n2+1−∑k=0∞1(2k+1)2+1 Each of the two sums can be evaluated by contour methods. In short, for a sufficient rational function f(z), limN→∞∑k=−Nk=Nf(k) is equal to the negative of the sum of the residues of πf(z)cot⁡(πz) at the poles of f(z). We can use this because we are summing an even function. This yields ∑n=0∞14n2+1=14(2+πcoth⁡(π2)) ∑k=0∞1(2k+1)2+1=14πtanh(π2) Subtract these (and simplify) to find ∑n=0∞(−1)xx2+1=12(πcsch(π)−1) Now we can conclude our first step by writing... ∑x=1∞(−1)x−x2−2x+1x4+2x2+1=−12(πcsch(π)−1)−∑x=1∞(−1)x2(x−1)(x2+1)2 It may be tempting to try the same approach for the remaining series but we are not summing an even function so it will not be as easy.

Palandriy0 · Accepted Answer

For the summation S=∑n=1∞(−1)n−n2−2n+1n4+2n2+1 gave a result: 18(2π2csch2(π2)+(1−i)(ψ(1)(12−i2)+i((−4+4i)+ψ(1)(12+i2)+ψ(1)(1−i2))+ψ(1)(1+i2))) which is approximately 0.31061370769015654201991515962234635408157816305055 (this could be computed for as many significant figures as required). so, T=∑n=0∞(−1)n2(n−1)(n2+1)2 then, (−18+i8)((2+2i)+(1+i)π(πcsch2(π2)+2csch(π))+ψ(1)(12−i2)+iψ(1)(12+i2)+iψ(1)(1−i2)+ψ(1)(1+i2))

I want to speed up the convergence of a series

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