How do you find the interval of convergence #Sigma (x+2)^n/(nsqrtn)# from #n=[1,oo)#?
1 Answer
Explanation:
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
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
So both -3 and -1 work. Therefore, the interval is:
#-3 le x le -1#
Final Answer