How do you find the interval of convergence #Sigma x^n/n^2# from #n=[1,oo)#?

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Answer 1 · 2018-07-05T05:56:05

The interval of convergence is #-1<=x<=1#

Use the Ratio test to compute the convergence interval

#a_n=x^n/n^2#

#|a_(n+1)/a_n|=|x^(n+1)/(n+1)^2*n^2/x^n|#

#=(n^2x)/(n+1)^2#

Calculate the limit as #n->oo#

Therefore,

#lim_(n->oo)|(n^2x)/(n+1)^2|#

#=|x|*lim_(n->oo)|(n^2)/(n+1)^2|#

#=|x|*1#

#=|x|#

The series converges for #|x|<1#

Now, check the convergence when, #|x|=1#

When #x=1#

#sum_1^oo1^n/n^2#

By the #" p-series test"#, #p=2# and #p>1#, the series converges

When #x=-1#

#sum_1^oo(-1)^n/n^2=sum_1^oo|(-1)^n/n^2|=sum_1^oo1/n^2#

By the #" p-series test"#, #p=2# and #p>1#, the series converges

The series converges absolutely