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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.

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    $\begingroup$ Possible duplicate of Does the infinite series $\sum_{n=1}^\infty \frac{\ln n}{n}$ converge? $\endgroup$ – Martin R Aug 25 at 5:14
  • $\begingroup$ Also: math.stackexchange.com/q/984993/42969, math.stackexchange.com/q/497299/42969 $\endgroup$ – Martin R Aug 25 at 5:15
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    $\begingroup$ What you have looks correct but, more simply, note for $n \ge 3$ that $\frac{\ln n}{n} \gt \frac{1}{n}$, and the harmonic series is a very well known divergent series. $\endgroup$ – John Omielan Aug 25 at 5:18
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    $\begingroup$ The title asks if the series converge (then it would have duplicates), but the body asks for verification of your solution (then not). Which one of those questions are you actually asking for? And if second, I suggest to change title to something like "Is my proof of X correct?" $\endgroup$ – Sil Aug 25 at 5:18
  • $\begingroup$ @Sil: Essentially the same method is used here and answered “This looks great.” So both questions are answered in the above listed duplicate targets. $\endgroup$ – Martin R Aug 25 at 5:22
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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.

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Your answer to this question is perfect .

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

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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.

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