How do you find the interval of convergence Sigma (-1)^nx^n/n from n=[1,oo)?

1 Answer
Mar 1, 2017

The series:

sum_(n=1)^oo (-1)^n x^n/n

is convergent for x in (-1,1].

Explanation:

Given the series:

sum_(n=1)^oo (-1)^n x^n/n

we can use the ratio test by evaluating:

abs(a_(n+1)/a_n) = abs ( ( (-1)^(n+1) x^(n+1) /(n+1) ) / ( (-1^n) x^n/n)) = n/(n+1) absx

So that the ratio limit is:

lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) n/(n+1) absx = abs x

meaning that for abs(x) < 1 the series is absolutely convergent and for abs(x) > 1 it is not convergent.

For abs x = 1 we have two cases:

(i) x = 1

sum_(n=1)^oo (-1)^n x^n/n = sum_(n=1)^oo (-1)^n/n which is the alternate harmonic series and is convergent.

(ii) x = -1

sum_(n=1)^oo (-1)^n x^n/n = sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n which is the harmonic series and is divergent.

In conclusion the series is convergent for x in (-1,1]