# How do you use the comparison test for sum (((ln n)^3) / (n^2)) n=1 to n=oo?

Jun 8, 2015

You've written:

${\sum}_{n = 1}^{\infty} \frac{{\ln}^{3} n}{n} ^ 2$

Let's call that ${\sum}_{n = 1}^{\infty} {a}_{n}$.

The Limit Comparison Test says (paraphrased):

For any two series that can be written as $\sum {a}_{n} \ge 0$ and $\sum {b}_{n} > 0$, define some limit $c$ as:

$c = {\lim}_{n \to \infty} \sum {a}_{n} / \left({b}_{n}\right)$

If the value of c turns out to be positive ($c > 0$) and finite, then the series ${a}_{n}$ converges. Otherwise, ${a}_{n}$ diverges.

There's an easy way to choose these series ${a}_{n}$ and ${b}_{n}$. We can choose a series ${b}_{n}$ that behaves just like ${a}_{n}$ by adding a $+ 1$.

Therefore, let:

${a}_{n} = \frac{{\ln}^{3} n}{n} ^ 2$

${b}_{n} = \frac{{\ln}^{3} \left(n + 1\right)}{n + 1} ^ 2$

So, now we get:

$c = {\lim}_{n \to \infty} \frac{{\ln}^{3} n}{n} ^ 2 \cdot {\left(n + 1\right)}^{2} / \left({\ln}^{3} \left(n + 1\right)\right)$

$= {\lim}_{n \to \infty} {\left(n + 1\right)}^{2} / {n}^{2} \frac{{\ln}^{3} n}{{\ln}^{3} \left(n + 1\right)}$

(Nono, don't use L'Hopital's Rule here; there's an easy way to do this.)

This limit can be split into a product of limits, like so:

$= {\lim}_{n \to \infty} {\left(n + 1\right)}^{2} / {n}^{2} \cdot {\lim}_{n \to \infty} \frac{{\ln}^{3} n}{{\ln}^{3} \left(n + 1\right)}$

Imagine, as $n \to \infty$, the $+ 1$ becomes really minor, so both limits approach the result of omitting the $+ 1$. The two limits are each $1$ because everything cancels out as $n \to \infty$. Therefore, the limit of the total result is $1$.

${\lim}_{n \to \infty} {a}_{n} / \left({b}_{n}\right) = 1 > 0$.

Therefore, the sum ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges.