Is my analysis of the series $\sum_{n=1}^\infty{\frac{\ln n}{n}}$ correct?

Question

I came across the following series and I'm supposed to analyse whether it converges or not. $$\sum_{n=1}^\infty{\frac{\ln n}{n}}$$

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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.

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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.

Answer 1

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.

Answer 2

Your answer to this question is perfect .

Answer 3

$$ \sum_{n = 1}^{N}\frac{\log n}{n} > \sum_{n = 1}^{N}\frac{1}{n} = \log N + \gamma + O(1/N) $$ Hence it is divergent by comparison with the harmonic series $1/n$.

Answer 4

I'm writing this answer just for the sake of providing a different answer, as well as supplying you with more series testing techniques aside from direct comparison and the integral test. Recall the following theorem:

Cauchy-Condensation Test: If the terms of a series $\sum_{n=1}^{\infty} a_n$ are nonnegative and decreasing monotonically to zero, then the series converges if and only if the related series $\sum_{k=1}^{\infty} 2^k a_{2^k}$ converges. So in your case, we need only check the convergence of

$$\sum_{k=1}^{\infty} 2^k \frac{\ln(2^k)}{2^k}$$

But notice that

$$\sum_{k=1}^{\infty} 2^k \frac{\ln(2^k)}{2^k} = \sum_{k=1}^{\infty} k \ln(2)$$

From here I hope it's easy to see that this series diverges.