Question #db1ed

We want to determine the convergence or divergence of

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

I agree with your conclusion that the limit comparison test is a good choice here.

The question then becomes, what should be chosen as #a_n# and #b_n#?

When doing the limit comparison test, it's often good to let #a_n# be the given sequence, so here let #a_n=sqrt(n+1)/(n^2-n)#.

The difficult part is choosing #b_n#. A good trick is to let #b_n# mimic the end behavior of whatever you've chosen for #a_n#.

For #sqrt(n+1)/(n^2-n)#, the dominating term as #n# goes to infinity is just #sqrtn#—that is, the constant becomes irrelevant. In the denominator, the #n^2# term grows faster than the #n# term, so the #n^2# term overpowers the #n#.

So, as #n# goes to infinity, we see that #sqrt(n+1)/(n^2-n)approxsqrtn/n^2=1/n^(3/2)#.

We then let #b_n=1/n^(3/2)#.

We can now proceed with the limit comparison test, which tells us to take #lim_(nrarroo)a_n/b_n#.

#lim_(nrarroo)a_n/b_n=lim_(nrarroo)(sqrt(n+1)/(n^2-n))/(1/n^(3/2))=lim_(nrarroo)(n^(3/2)sqrt(n+1))/(n^2-n)#

You may recognize that since the degrees of the numerator and denominator are equal, the limit will be the ratio of the coefficients, or #1/1=1#. If you're not comfortable with this, we can still find the limit algebraically.

#lim_(nrarroo)a_n/b_n=lim_(nrarroo)(sqrtnsqrt(n+1))/(n-1)=lim_(nrarroo)(sqrtnsqrtnsqrt(1+1/n))/(n(1-1/n))#

#color(white)(lim_(nrarroo)a_n/b_n)=lim_(nrarroo)sqrt(1+1/n)/(1-1/n)#

As #n# approaches infinity, #1/n# approaches #0#.

#lim_(nrarroo)a_n/b_n=sqrt(1+0)/(1-0)=1#

We see that #lim_(nrarroo)a_n/b_n# is real, positive, and defined. The limit comparison test tells us that if this indeed is the case, then:

• #suma_n# and #sumb_n# both converge or
• #suma_n# and #sumb_n# both diverge

We don't yet know whether or not #suma_n# diverges--that's what we're trying to figure out. We do, however, know the convergence/divergence of #sumb_n#.

Since #b_n=1/n^(3/2)#, we see that #sumb_n=sum_(n=2)^oo1/n^(3/2)# converges through the p-series test (that is, since #3/2gt1#).

Since we know that #sumb_n# converges, we can conclude that #suma_n=sum_(n=2)^oosqrt(n+1)/(n(n-1))# converges as well.