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

Question

Charles Arocha · Accepted Answer

The series:

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

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

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]#

Ezekiel Alexander · Answer

undefined The interval of convergence for the series $ \sum_{n=1}^{\infty} (-1)^n\frac{x^n}{n} $ can be found using the ratio test. The ratio test states that the series converges if the limit as $ n $ approaches infinity of the absolute value of $ \frac{a_{n+1}}{a_n} $ is less than 1. In this case, $ a_n = (-1)^n\frac{x^n}{n} $. Applying the ratio test, you will find that the interval of convergence is $ -1 < x \leq 1 $, including $ x = -1 $ and excluding $ x = 1 $.