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It seems that

 

Limit(N+(sum((-1)^n*Sum(1/n^x, x = 1 .. N), n = 1 .. infinity)), N = infinity)=log(2)

 evalf(300+sum((-1)^n*(Sum(1/n^x, x = 1 .. 300)), n = 1 .. infinity), 30)

gives

0.693147180559945309417232121.

 sum(1/n^x, x = 1 .. infinity)

gives

1/(n-1).
limit

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