How do you find the interval of convergence of #Sigma (x+10)^n/(lnn)# from #n=[2,oo)#?

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Answer 1 · 2017-04-11T22:40:14

The series:

#sum_(n=0)^oo (x+10)^n/lnn#

is convergent for # x in [-11,9)# and absolutely convergent in the interior of the interval.

(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)#