∑ k = − ∞ ∞ 1 0 ( 1 0 0 k ) 2 1 = ϑ 3 ( 0 , 1 0 0 0 0 1 0 1 ) . You may use whichever expression that you wish. The answer to the problem is the same. The problem will be stated using the first form.
What is the numerical approximation to lo g 1 0 ( ∣ ∣ ∣ ∣ ( ∑ k = − ∞ ∞ 1 0 ( 1 0 0 k ) 2 1 ) − 1 0 0 lo g ( 1 0 ) π ∣ ∣ ∣ ∣ ) .
As the problem was computed using Wolfram Mathematica 12.1, no solution will be given by the author.
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