How do you determine whether the sequence a_n=n/(ln(n)^2 converges, if so how do you find the limit?

Jan 25, 2017

The series:

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

is divergent.

Explanation:

Note: The question is not mentioning it expressly, but in the answer, I assume the series starts from $n = 2$ as ${a}_{n}$ is undefined for $n = 1$

A necessary condition for any series to converge is Cauchy's condition that:

${\lim}_{n \to \infty} {a}_{n} = 0$

To verify whether this condition is met, first we note that

${a}_{n} = f \left(n\right)$

where $f \left(x\right) = \frac{x}{\ln} {\left(x\right)}^{2}$ is a continuous function for $x \in \left(1 , + \infty\right)$

So if the limit ${\lim}_{x \to \infty} f \left(x\right)$ exixst it must be the same as ${\lim}_{n \to \infty} {a}_{n}$

Now:

${\lim}_{x \to \infty} f \left(x\right) = {\lim}_{x \to \infty} \frac{x}{\ln x} ^ 2$

is the inderminate form $\frac{\infty}{\infty}$ and we can solve it using the l'Hospital's rule (twice):

${\lim}_{x \to \infty} \frac{x}{\ln x} ^ 2 = {\lim}_{x \to \infty} \frac{1}{\frac{2 \ln x}{x}} = {\lim}_{x \to \infty} \frac{x}{2 \ln x} = {\lim}_{x \to \infty} \frac{1}{\frac{2}{x}} = {\lim}_{x \to \infty} \frac{x}{2} = + \infty$

Thus also:

${\lim}_{n \to \infty} \frac{n}{\ln n} ^ 2 = + \infty$

which implies that the series is not convergent.

As it has positive terms, the series is then divergent.