# Question #db1ed

##### 1 Answer

Yes—through the limit comparison test, you can conclude that

#### Explanation:

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

When doing the limit comparison test, it's often good to let

The difficult part is choosing

For

So, as

We then let

We can now proceed with the limit comparison test, which tells us to take

#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

#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

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

We see that *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

Since

Since we know that