Question

Question #db1ed

Answer 1

Yes—through the limit comparison test, you can conclude that `sum_(n=2)^oosqrt(n+1)/(n(n-1))` converges.

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 `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.