For what values of a&gt;0 is the equation∑n=1∞2(n−0.5)2+a=πatrue ?

Question

For what values of a&amp;gt;0 is the equation∑n=1∞2(n−0.5)2+a=πatrue ?

mastifo5h · Accepted Answer

We start from the result linked in the question:∑n=−∞∞1(n+α)2+γ2=πγsinh2γπcosh2γπ−cos⁡2απ.For the specific case of α=−12, it is not too hard to see that∑n=−∞01(n−12)2+γ2=∑n=1∞1(n−12)2+γ2and so we have2∑n=1∞1(n−12)2+γ2=∑n=−∞∞1(n−12)2+γ2=πγsinh2γπcosh2γπ−cos⁡π=πγsinh2γπcosh2γπ+1.But sinh2x=2sinhxcoshx and cosh2x+1=2cosh2x, so∑n=1∞2(n−12)2+γ2=πγtanhγπ=πatanh(aπ),where we have identified γ=a in the last step. As the OP notes, this should have a finite limit as a→0, and a bit of work shows that this limit is indeed finite and equal to π2Thus, the given formula is true for all values of a for which tanh(aπ)=1, which is... none of them. For all real x,−1&amp;lt;tanh(x)&amp;lt;1. However, the hyperbolic tangent function does asymptotically approach 1; in fact, asymptotically we havetanh(aπ)≈1−2e−2aπEven for a=1, we have tanh(π)≈0.996 and so the result holds to within less than 1%. It is not too hard to envision an amateur numerologist stumbling upon this relation &quot;experimentally&quot; with a computer and thinking that it must be true. It is a remarkably good approximation, particularly for large values of a, but it is not in fact true.

Zernerqcw · Accepted Answer

Consider the partial sumSp=∑n=1p2(n−12)2+a=∑n=1p2(n+12(−2−a−1))(n+12(2−a−1))Now, using the asymptoticsψ(n)=log⁡(n)−12n−112n2+O(1n4)Using it twice and continuing with Taylor seriesSp=ψ(12+−a)−ψ(12−−a)−a−2p+4a+16p3+O(1p5)and using nowψ(12+b)−ψ(12−b)=πtan⁡(πb)thusψ(12+−a)−ψ(12−−a)−a=πatanh(πa)This is what Wolfram Alpha provided.

For what values of a>0 is the equation \sum_{n=1}^{\infty} \frac{2}{(n-0.5)^2+a} =

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