# How do you show whether sum_(n=2)^oo 1/ln^3(n) converges or diverges?

Dec 4, 2016

${\sum}_{n = 2}^{\infty} \frac{1}{\ln} ^ 3 \left(n\right)$ diverges by the Cauchy condensation test

#### Explanation:

The Cauchy condensation test states that if $f \left(n\right)$ is nonnegative and nonincreasing real sequence, then ${\sum}_{n = 1}^{\infty} f \left(n\right)$ converges if and only if the sum ${\sum}_{n = 0}^{\infty} {2}^{n} f \left({2}^{n}\right)$ converges as well.

Let $f \left(n\right) = \left\{\begin{matrix}\frac{1}{\ln} ^ 3 \left(n\right) \mathmr{if} n > 1 \\ f \left(n + 1\right) \mathmr{if} n = 1\end{matrix}\right.$

As adding a finite value to a series does not change whether it converges or diverges, we can add $\frac{1}{\ln} ^ 3 \left(2\right)$ to the given series without changing the result. Doing so, we can apply the Cauchy condensation test:

$\frac{1}{\ln} ^ 3 \left(2\right) + {\sum}_{n = 2}^{\infty} \frac{1}{\ln} ^ 3 \left(n\right) = {\sum}_{n = 1}^{\infty} f \left(n\right)$ converges if and only if ${\sum}_{n = 0}^{\infty} {2}^{n} f \left({2}^{n}\right) = {\sum}_{n = 1}^{\infty} {2}^{n} / {\ln}^{3} \left({2}^{n}\right)$ converges.

Now, looking at the condensed sum, we have

${\sum}_{n = 1}^{\infty} {2}^{n} / {\ln}^{3} \left({2}^{n}\right) = {\sum}_{n = 1}^{\infty} {2}^{n} / {\left(n \ln \left(2\right)\right)}^{3} = {\sum}_{n = 1}^{\infty} {\ln}^{- 3} \left(2\right) {2}^{n} / {n}^{3}$

which diverges by the divergence test, as ${\lim}_{n \to \infty} | {2}^{n} / {n}^{3} | = \infty$.

As ${\sum}_{n = 0}^{\infty} {2}^{n} f \left({2}^{n}\right)$ diverges, so does ${\sum}_{n = 1}^{\infty} f \left(n\right) = \frac{1}{\ln} ^ 3 \left(2\right) + {\sum}_{n = 2}^{\infty} \frac{1}{\ln} ^ 3 \left(n\right)$, and thus so does the given series.

Dec 4, 2016

My first intuition is to show that $\frac{1}{\ln} ^ 3 \left(n\right) > \frac{1}{n}$ to claim that ${\sum}_{n = 2}^{\infty} \frac{1}{\ln} ^ 3 \left(n\right)$ diverges through direct comparison, since ${\sum}_{n = 2}^{\infty} \frac{1}{n}$ diverges as the harmonic series.

Notice that $\frac{1}{\ln} ^ 3 \left(n\right) = \frac{1}{n} \left(\frac{n}{\ln} ^ 3 \left(n\right)\right)$. From this we see that $\frac{1}{\ln} ^ 3 \left(n\right) > \frac{1}{n}$ if $\frac{n}{\ln} ^ 3 \left(n\right) > 1$.

To see if this is true as these series extend infinitely, we can take the infinite limit of $\frac{n}{\ln} ^ 3 \left(n\right)$.

${\lim}_{n \rightarrow \infty} \frac{n}{\ln} ^ 3 \left(n\right)$

This is indeterminate in the form $\frac{\infty}{\infty}$, so we can apply L'Hospital's rule and tale the derivative of the numerator and denominator separately.

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

Reapplying L'Hospitals (we can see a pattern forming):

$= {\lim}_{n \rightarrow \infty} \frac{1}{6 \ln \left(n\right) \frac{1}{n}} = {\lim}_{n \rightarrow \infty} \frac{n}{6 \ln \left(n\right)}$

L'Hospital's once more:

$= {\lim}_{n \rightarrow \infty} \frac{1}{6 \frac{1}{n}} = {\lim}_{n \rightarrow \infty} \frac{n}{6} = \infty$

Thus we've seen that $\frac{n}{\ln} ^ 3 \left(n\right)$ will be far greater than $1$, and that we can state that $\frac{1}{\ln} ^ 3 \left(n\right) = \frac{1}{n} \left(\frac{n}{\ln} ^ 3 \left(n\right)\right) > \frac{1}{n}$ for sufficiently large values of $n$.

Through direct comparison, ${\sum}_{n = 2}^{\infty} \frac{1}{\ln} ^ 3 \left(n\right)$ diverges.