# Question #db1ed

Dec 29, 2016

Yes—through the limit comparison test, you can conclude that ${\sum}_{n = 2}^{\infty} \frac{\sqrt{n + 1}}{n \left(n - 1\right)}$ converges.

#### Explanation:

We want to determine the convergence or divergence of

${\sum}_{n = 2}^{\infty} \frac{\sqrt{n + 1}}{n \left(n - 1\right)} = {\sum}_{n = 2}^{\infty} \frac{\sqrt{n + 1}}{{n}^{2} - n}$

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 ${a}_{n} = \frac{\sqrt{n + 1}}{{n}^{2} - n}$.

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 $\frac{\sqrt{n + 1}}{{n}^{2} - n}$, the dominating term as $n$ goes to infinity is just $\sqrt{n}$—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 $\frac{\sqrt{n + 1}}{{n}^{2} - n} \approx \frac{\sqrt{n}}{n} ^ 2 = \frac{1}{n} ^ \left(\frac{3}{2}\right)$.

We then let ${b}_{n} = \frac{1}{n} ^ \left(\frac{3}{2}\right)$.

We can now proceed with the limit comparison test, which tells us to take ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n}$.

${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = {\lim}_{n \rightarrow \infty} \frac{\frac{\sqrt{n + 1}}{{n}^{2} - n}}{\frac{1}{n} ^ \left(\frac{3}{2}\right)} = {\lim}_{n \rightarrow \infty} \frac{{n}^{\frac{3}{2}} \sqrt{n + 1}}{{n}^{2} - n}$

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

${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = {\lim}_{n \rightarrow \infty} \frac{\sqrt{n} \sqrt{n + 1}}{n - 1} = {\lim}_{n \rightarrow \infty} \frac{\sqrt{n} \sqrt{n} \sqrt{1 + \frac{1}{n}}}{n \left(1 - \frac{1}{n}\right)}$

$\textcolor{w h i t e}{{\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n}} = {\lim}_{n \rightarrow \infty} \frac{\sqrt{1 + \frac{1}{n}}}{1 - \frac{1}{n}}$

As $n$ approaches infinity, $\frac{1}{n}$ approaches $0$.

${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = \frac{\sqrt{1 + 0}}{1 - 0} = 1$

We see that ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n}$ is real, positive, and defined. The limit comparison test tells us that if this indeed is the case, then:

• $\sum {a}_{n}$ and $\sum {b}_{n}$ both converge or
• $\sum {a}_{n}$ and $\sum {b}_{n}$ both diverge

We don't yet know whether or not $\sum {a}_{n}$ diverges--that's what we're trying to figure out. We do, however, know the convergence/divergence of $\sum {b}_{n}$.

Since ${b}_{n} = \frac{1}{n} ^ \left(\frac{3}{2}\right)$, we see that $\sum {b}_{n} = {\sum}_{n = 2}^{\infty} \frac{1}{n} ^ \left(\frac{3}{2}\right)$ converges through the p-series test (that is, since $\frac{3}{2} > 1$).

Since we know that $\sum {b}_{n}$ converges, we can conclude that $\sum {a}_{n} = {\sum}_{n = 2}^{\infty} \frac{\sqrt{n + 1}}{n \left(n - 1\right)}$ converges as well.