# How do you use the integral test to determine if ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+... is convergent or divergent?

Jan 3, 2017

It is divergent. See explanation.

#### Explanation:

First write the series:

$\ln \frac{2}{2} + \ln \frac{3}{3} + \ln \frac{4}{4} + \ldots = {\sum}_{n = 2}^{\infty} \ln \frac{n}{n}$

Before getting into the integral test, we must assure two things first: for the integral test to apply to ${\sum}_{n = N}^{\infty} f \left(n\right)$ we must have ${a}_{n} > 0$ for the given interval and $f \left(n\right)$ must be decreasing on the same interval.

Both of these are true since $\ln \left(n\right) > 0$ and $n > 0$ on $n \in \left[2 , \infty\right)$, so $\ln \frac{n}{n} > 0$ on the same interval.

Furthermore, $\ln \left(n\right)$ grows slower than $n$ so we see that $\ln \frac{n}{n}$ is decreasing because $n$ overpowers $\ln \left(n\right)$ in the numerator. You can also show this by taking the derivative of $\ln \frac{n}{n}$ and showing it's always negative on $n \in \left[2 , \infty\right)$.

So, we see the integral test applies. The integral test states that if the two aforementioned conditions are met, then for ${\sum}_{n = N}^{\infty} f \left(n\right)$, evaluate the improper integral ${\int}_{N}^{\infty} f \left(x\right) \mathrm{dx}$.

If the integral converges to a real, finite value, then the series converges. If the integral diverges, then the series does too.

So, we take the integral ${\int}_{2}^{\infty} \ln \frac{x}{x} \mathrm{dx}$. Take the limit as it goes to infinity:

${\int}_{2}^{\infty} \ln \frac{x}{x} \mathrm{dx} = {\lim}_{b \rightarrow \infty} {\int}_{2}^{b} \ln \frac{x}{x} \mathrm{dx}$

Letting $u = \ln \left(x\right)$, so $\mathrm{du} = \frac{1}{x} \mathrm{dx}$:

$= {\lim}_{b \rightarrow \infty} {\int}_{\ln} {\left(2\right)}^{\ln} \left(b\right) u \textcolor{w h i t e}{.} \mathrm{du}$

$= {\lim}_{b \rightarrow \infty} {\left[\frac{1}{2} {u}^{2}\right]}_{\ln} {\left(2\right)}^{\ln} \left(b\right)$

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

As $b \rightarrow \infty$, we see that $\ln \left(b\right) \rightarrow \infty$, so:

$= \infty$

The integral diverges. Thus, we see that ${\sum}_{n = 2}^{\infty} \ln \frac{n}{n}$ diverges as well.