I came across the following series and I'm supposed to analyse whether it converges or not.
$$\sum_{n=1}^\infty{\frac{\ln n}{n}}$$
My attempt:
At first sight, the thought of using the integral test came to my mind as each term of the series would be non-negative. But as I proceeded, I noticed that the function $f(x)=\frac{\ln x}{x}$ is not a monotonic function on the domain $x \in [1, \infty), \, \forall x \in \mathbb{R}$ as:
$$f'(x)=\frac{1-\ln x}{x^2}$$
Since $f'(x)<0, \, \forall \,x>e$, therefore I started analysing $\sum_{n=3}^\infty\frac{\ln x}{x}$ for which $f(x)$ is monotonic and decreasing. I solved it as follows:
$$\sum_{n=1}^\infty\frac{\ln x}{x}=\frac{\ln1}{1}+\frac{\ln2}{2}+\sum_{n=3}^\infty\frac{\ln x}{x}$$
For $\sum_{n=1}^\infty\frac{\ln x}{x}$ to converge, $\int_3^\infty{f(x)}dx$ must also converge which would eventually lead to the convergence of $\sum_{n=1}^\infty\frac{\ln x}{x}$ as the initial two terms of the series are constants.
$$\int_3^\infty{f(x)}dx=\int_3^\infty\frac{\ln x}{x}dx=\int_3^\infty{\ln{x}\,d(\ln x)}=\infty$$
Since the integral of $f(x)$ diverges, therefore the corresponding sum must also diverge, this implies that the series $\sum_{n=1}^{\infty}\frac{\ln n}{n}$ must also diverge.
I am unsure if my analysis is correct or not. It would be helpful if some person suggests a better method or points out some mistakes in my attempt, if any.
Best Answer
Although your method is correct and works, the method to determine convergence that seems most obvious to me (and the easiest to do) is the direct comparison test with $\frac{1}{n}$. Specifically $$\frac{\ln(n)}{n} > \frac{1}{n}$$ for $n \ge 3$. Therefore $$\sum_{n=1}^\infty \frac{\ln(n)}{n} > \sum_{n=1}^\infty \frac{1}{n}$$ which is known to diverge.