(i) For every #n > 2# we have #ln n < n#, so if # ( x+10 ) >= 1#:
# (x+10)^n/ln n > 1/n#
and the series is divergent.
(ii) For #abs(x+10) < 1# we have:
#abs(x+10)^n/ln n < abs(x+10)^n#
Since #sum_(n=0)^oo abs(x+10)^n# is a geometric series of ratio #r < 1#, it is convergent and then also:
#sum_(n=0)^oo (x+10)^n/lnn# is absolutely convergent.
(iii) For #(x+10) <= -1 # we can write the series as an alternating series:
#sum_(n=0)^oo (-1)^n abs(x+10)^n/lnn#
and apply the Leibniz test:
Clearly for #(x+10) < -1# we have
#lim_(n->oo) a_n = lim_(n->oo) abs(x+10)^n/lnn = oo #
and the series is not convergent, while for #(x+10) = -1#
#lim_(n->oo) a_n = lim_(n->oo) 1/lnn = 0 #
and
#a_(n+1)/a_n = lnn/ln(n+1)<1#
so the series is convergent.
In conclusion the series is convergent for:
#-1 <= x+10 < 1#
That is for # x in [-11,9)#
