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Answer 1 · 2017-09-26T10:41:30

# -5 < x < -1 #

#((x + 3)^(n-1))/(2^n) = ((x + 3)^n)/((x+3)*2^n) = 1/(x+3) ((x+3)/2)^n#

Then we get that #sum ((x+3)/2)^n# converges when #|\ (x+3)/2 | < 1 #.

So we solve for #x#

# -1 < (x+3)/2 < 1 #

# -2 < x+3 < 2 #

# -5 < x < -1 #

Then we have that # 1/(x+3) # is just a constant, and so we get that

# 1/(x+3) sum ((x+3)/2)^n# converges when # -5 < x < -1 #.