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Question: evalf/Sum and epsilon

Question:evalf/Sum and epsilon

_Maxim_ 729
sum digits numeric
November 19 2017
1

This remains unevaluated:

(evalf@Sum)((1-10^(-3))^k/k, k = 1 .. infinity)

This might (or might not) be related to the fact that evalf/Sum doesn't have a tolerance (epsilon) setting. If I want to evaluate the sum with 1-10^-20 instead of 1-10^-3 directly, I need to increase Digits. But then the tolerance setting will increase too to about 10^(-Digits+1), and evalf/Sum will return unevaluated again because it won't be able to attain the requested accuracy. Whereas often only a few digits are needed -- e.g., to see how the sum grows with n -- and a tolerance value of 10^-3 or so would be sufficient.
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