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# Precision of numerical series value in Wolfram|Alpha

Posted 1 year ago
 Error It is gives $$\sum_{k=0}^{\infty}\frac{(-1)^k}{k^2+2k+2}\approx 0.363998$$ but exact value: $$\sum_{k=0}^{\infty}\frac{(-1)^k}{k^2+2k+2}=\frac{1}{2}-\frac{\pi}{e^{\pi}-e^{-\pi}}\approx 0.3639854725089334$$
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Posted 1 year ago
 With Mathematica 13.1,numerically:  NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 32] (*0.36398547250893341852488170865923*) NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}, Method -> "WynnEpsilon", NSumTerms -> 10^5, WorkingPrecision -> 32] (*0.36398547250893341852488170816398*) Symbolically: Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}] (*1/2 (1 - \[Pi] Csch[\[Pi]])*) N[%, 32] (*0.36398547250893341852488170865923*) If you use WolframAlfa answer is not precisie ,because for quick answer to you from web. Try to put: N[Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}] ,32] Give you: 0.36398547250893341852488170816398
Posted 1 year ago
 Thanks. Yes, I used Wolframa|Alpha. But I think that it is not a good idea to have more velocity and error results from web form.