What is the interval of convergence of #sum ((x − 4)^n)/(n*(−9)^n)#?

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Answer 1 · 2016-07-03T09:19:56

#(-5, 13]#

Compare with # ln(1+X)=sum(-1)^(n-1)X^n/n, n=1,, 2, 3, ..#,

# -1 < X <=1#.

Easily, #X =(x-4)/9#.

So,# -1< (x-4)/9 <=1#.

And so, explicitly, #-5 < x <=13#.

Answer 2 · 2016-07-03T09:26:20

#sum_{n=1}^{oo} ((x − 4)^n)/(n(−9)^n) =-log_e abs((5 + x)/9)#

is convergent for #-5 < x < 13#

#sum_{n=1}^{oo} ((x − 4)^n)/(n(−9)^n) =sum_{n=0}^{oo}(((x-4)/(-9))^n)/n #

For #abs y < 1#

#d/(dy)sum_{k=1}^{oo}(y^k)/k = sum_{k=1}^{oo}y^{k-1}=1/(1-y)#

so

#sum_{k=1}^{oo}(y^k)/k=-log_e abs(1-y)#

then for #abs((x-4)/(-9))<1 -> -5 < x <13# the series is convergent