What is the interval of convergence of #sum (-1)^n * (x^n/n) #?

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Answer 1 · 2017-11-11T20:12:19

See below.

Calling #S(x) = sum_(k=0)^oo (-1)^k x^k/k# we have

#(dS)/(dx) = sum_(k=1)^oo(-1)^k x^(k-1) = 1/x sum_(k=1)^oo(-1)^k x^k#

now assuming #absx < 1#

#(dS)/(dx) = x(1/(x+1)-1) = -x^2/(x+1)#

and integrating

#S(x) = 2 (1 + x) - 1/2 (1 + x)^2 - Log(1 + x)+C_0#

but #S(0) = 1 rArr C_0 = -1/2#

then

#S(x) = 2 (1 + x) - 1/2 (1 + x)^2 - Log(1 + x)-1/2#

which converges for #abs x < 1#