Using the integral test, how do you show whether #sum 1/(nln(3n))# diverges or converges from n=1 to infinity?

Question

Christopher Allers · Accepted Answer

Diverges.

Let #f(x)=1/(xln3x)#. Now, the integral test tells us that if #f(x)# is positive, continuous, and decreasing on #[k, oo)#, the convergence or divergence of #int_k^oof(x)dx# will tell us about the convergence or divergence of #sum_(n=k)^oof(n)#. So, we have #sum_(n=1)^oo1/(nln3n)# and we will be evaluating #int_1^oo1/(xln3x)dx#. First, let's find the general antiderivative for #intdx/(xln3x).# Let #u=ln(3x), du=(3/(3x))dx=dx/x# So, we then have #int(du)/u=ln|u|=ln|ln(3x)|# Now, #int_1^oodx/(xln3x)=lim_(t->oo)int_1^tdx/(xln3x)=lim_(t->oo)(ln|ln3x|)|_1^t# Evaluate the limit: #lim_(t->oo)(ln(ln3t)-ln(ln3))=oo-ln(ln3)# We can drop the absolute value bars. Not working with any negatives. The integral diverges; therefore, so does the series.

Isabella Ackerman · Answer

undefined To determine whether the series $\sum_{n=1}^{\infty} \frac{1}{n\ln(3n)}$ converges or diverges using the integral test, we need to analyze the convergence of the corresponding improper integral.

The integral test states that if $f(n)$ is a continuous, positive, and decreasing function for all $n$ greater than some positive integer $N$, and if the series $\sum_{n=N}^{\infty} f(n)$ and the integral $\int_{N}^{\infty} f(x) \, dx$ have the same convergence behavior, then the series either both converge or both diverge.

Let's consider the function $f(x) = \frac{1}{x\ln(3x)}$ and integrate it from $N$ to infinity:

$$\int_{N}^{\infty} \frac{1}{x\ln(3x)} \, dx$$

We'll evaluate this integral and check its convergence. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

We first integrate $f(x)$ using substitution. Let $u = \ln(3x)$, then $du = \frac{3}{x} \, dx$, which gives us:

$$\int \frac{1}{x\ln(3x)} \, dx = \int \frac{1}{u} \, du = \ln|u| + C = \ln|\ln(3x)| + C$$

Now, evaluating the improper integral from $N$ to infinity:

$$\lim_{b \to \infty} \int_{N}^{b} \frac{1}{x\ln(3x)} \, dx = \lim_{b \to \infty} \left[\ln|\ln(3x)|\right]_{N}^{b}$$

$$= \lim_{b \to \infty} \left[\ln|\ln(3b)| - \ln|\ln(3N)|\right]$$

Since $\ln|\ln(3b)|$ approaches infinity as $b$ approaches infinity, and $\ln|\ln(3N)|$ is a constant, the limit diverges.

Therefore, by the integral test, since the corresponding integral diverges, the series $\sum_{n=1}^{\infty} \frac{1}{n\ln(3n)}$ also diverges.