How do we prove (intuitively is fine) that $\sum_{n=1}^{\infty} \frac{\ln (n)}{n}$ diverges?

Question

From my Khan acad knowledge, we know that if $p>1$ it converges, otherwise it diverges. Intuitively we can tell that $\frac{\ln (n)}{n} > \frac{1}{n}$ whenever $n >e$. To further argue that it should converge I tried graphing it to show that the series indeed cover a larger area than a harmonic series ($\frac{1}{n}$), which should in my head result in a convergence if anything:

and yet, khan acad answer key concludes that "Since $\sum_{n=1}^{\infty} \frac{\ln (n)}{n} > \sum_{n=1}^{\infty} \frac{1}{n}$ "diverges".

I even tried pulling up the $p$-series "rules", while it's not a power that is larger than $1$, it is a value larger than $\frac{1}{n}$ for sure.

What am I missing here? Is it because I shouldn't be using graphs, if so how do we prove it?... I'm unable to understand how they concluded "Since $\sum_{n=1}^{\infty} \frac{\ln (n)}{n} > \sum_{n=1}^{\infty} \frac{1}{n}$ it "diverges".

Here is their answer key:

Answer 1

How do we prove (intuitively is fine) that $\sum_{n=1}^{\infty} > \frac{\ln (n)}{n}$ diverges?

I'm unable to understand how they concluded "Since $\sum_{n=1}^{\infty} \frac{\ln (n)}{n} > \sum_{n=1}^{\infty} > \frac{1}{n}$ it "diverges".

The course assumes that by this point one is aware that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges,so that any other series $a_k$ that $\sum_{n=1}^{\infty} a_n > \sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.

What can be more intuitive than $b_k$ is a divergent series, $a_k>b_k$ from some $k$ onwards implies that $b_k$ diverges.

Answer 2

Integral test. If $f:[1,\infty)\to [0,\infty)$ is continuous and if $f(x)$ is decreasing for all sufficiently large $x\in [1,\infty)$ then $\sum_{j=1}^nf(j)$ converges as $n\to\infty$ iff $\int_1^nf(x)dx$ converges as $n\to\infty$.

Proof: Take $2\le n_0\in\Bbb N$ such that $f(x)$ is decreasing for $x\ge n_0.$ For $n_1,n_2\in \Bbb N$ with $n_0\le n_1\le n_2$ we have $$0\le \int_{-1+n_1}^{n_2}f(x)dx\le \sum_{j=n_1}^{n_2}f(j)\le \int_{n_1}^{1+n_2}f(x)dx\le \sum_{j=-1+n_1}^{n_2}f(j).... QED.$$ For $f(x)=(\ln x)/x$ we have $f'(x)=(1-\ln x)/x^2<0$ for $x>e$. And for $n\ge 1$ we have $$\int_1^n f(x)dx=\int_1^n\frac {\ln x}{x}dx= \int_1^n(\ln x)d(\ln x)=\frac {1}{2}(\ln n)^2.$$

Answer 3

For instance by creative telescoping. You may easily realize that $\frac{\log x}{x}$ is half the derivative of $\log^2 x$, and

$$ \log^2(n+1)-\log^2(n) = (\log(n+1)+\log(n))\log\left(1+\frac{1}{n}\right)< 2\frac{\log(n+1)}{n}< 2\left(\frac{\log n}{n}+\frac{1}{n^2}\right). $$

This leads to

$$ \sum_{k=1}^{N}\frac{\log n}{n} > \frac{1}{2}\sum_{k=1}^{n}\left(\log^2(n+1)-\log^2(n)\right)-\sum_{k=1}^{N}\frac{1}{k^2}>\frac{\log^2(N+1)}{2}-\frac{\pi^2}{6}. $$

Answer 4

If the sum of f(n) diverges, f(n) >=0, and g(n) >=f(n) then g(n) diverges. And that’s the case here.

Intuitive divergence of 1/n: Sum for n= 1 is 1. Sum for 2<=n<= 2 is 1/2. Sum for 3<=n<=4 is 2 values >= 1/4, so >= 1/2. Sum for 5 <= n <= 8 is 4 values >= 1/8, so >= 1/2. Sum for 9 <= n <= 16 is 8 values >= 1/16, so >= 1/2. You keep adding up 1/2 so the sum diverges.

Intuitive divergence of ln n/ n: we have ln 3 >= ln 2, ln 5 >= 2 ln 2, ln 9 >= 3 ln 2, ln 17 >= 4 ln 2, so doing Ehe sum sums we add 0 + 1/2 + 1/2 + 2/2 + 3/2 + 4/2 etc, which diverges even faster.