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

Use the rule of the ratio test:

#lim_(n->oo) |a_(n+1)/a_n|<1#

Plug in the nth and (n+1)th term and solve for x.

#lim_(n->oo) |((x+2)^(n+1)/((n+1)sqrt(n+1)))/((x+2)^n/(nsqrtn))|<1#

#lim_(n->oo) |(x+2)^(n+1)/(x+2)^n * (nsqrtn)/((n+1)sqrt(n+1))| <1#

#lim_(n->oo) |(x+2)^(n+1)/(x+2)^n |* lim_(n->oo)|(nsqrtn)/((n+1)sqrt(n+1))| <1#

As n goes to infinity, n and n+1 will get infinitely closer together, so the second limit will approach 1.

#lim_(n->oo)abs(x+2)*1 < 1#

#abs(x+2) < 1#

#-1 < x+2 < 1#

#-3 < x < -1#

All we have to do now is figure out whether or not the endpoints work. Let's start with -3:

#sum_(n=1)^oo (-3+2)^n/(nsqrtn) = sum_(n=1)^oo (-1)^n/n^1.5#

This creates an alternating series. Since the bottom number will always increase, the absolute value of the whole fraction will always decrease, so the alternating series test is satisfied. Therefore #x=-3# is a solution.

Now for -1:

#sum_(n=1)^oo (-1+2)^n/n^1.5 = sum_(n=1)^oo 1/n^1.5#

This converges by the p-series test, since #p=1.5>1#.

So both -3 and -1 work. Therefore, the interval is:

#-3 le x le -1#

Final Answer