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

1 Answer
Jan 24, 2017

The series:

#sum_(n=1)^oo (-1)^n (x+2)^n/n#

is absolutely convergent for #x in (-3,-1)# and simply convergent for #x =-1#

Explanation:

To find the interval of convergence we can apply the ratio test, stating that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that:

#L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1#

If #L < 1 # the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

#abs (a_(n+1)/a_n) = abs (frac ( (x+2)^(n+1)/(n+1)) ((x+2)^n/n)) = abs(x+2) n/(n+1) #

so that:

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x+2)#

we can therefore conclude that the series is absolutely convergent for #abs (x+2) < 1# and divergent for #abs (x+2) > 1#.

In the case where #abs (x+2) = 1# we know that the test is inconclusive but we can see that for #x=-3# the series becomes:

#sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n#

that is divergent, while for #x= -1# we have:

#sum_(n=1)^oo (-1)^n/n#

that is convergent, but not absolutely convergent.

In conclusion the series converges for #x in (-3,-1]#