Asymptotic behaviour of $\displaystyle\sum_{n \leq x} \frac{\log(n)}{n}$

Question

This is my first question, I hope I won't make too many mistakes.

I have been given this question as an exercise, but I am struggling to find the solution. I have to calculate the asymptotic behaviour of $\displaystyle\sum_{n \leq x} \frac{\log(n)}{n}$. I have tried to apply the partial sums formula, which is:

Let $\phi \in C^1(0,\infty)$, $\{a_n\} \subset \mathbb{C}$,

$\displaystyle\sum_{n \leq x} a_n\phi(n) = A(x)\phi(x) - \int_1^x A(u)\phi'(u)\,du$, where $A(x) = \displaystyle\sum_{n \leq x} a_n$.

I have decided to put $a_n = 1$, and $\frac{\log(n)}{n} = \phi(n)$. In this way, I obtain

$\displaystyle\sum_{n \leq x}\frac{\log(n)}{n} = \left[x\right]\frac{\log(x)}{x} - \int_1^x \frac{(1 - \log(u))}{u^2} \left[u\right] du$

$= \log(x)-\frac{\left\{x\right\}\log(x)}{x}-\displaystyle\int_1^x \frac{u-\left\{u\right\}}{u^2} du + \displaystyle\int_1^x \frac{\log(u)(u-\left\{u\right\})}{u^2} du$

$=\log(x)-\frac{\left\{x\right\}\log(x)}{x}-\displaystyle\int_1^x \frac{1}{u} du + \displaystyle\int_1^x \frac{\left\{u\right\}}{u^2} du+ \displaystyle\int_1^x \frac{\log(u)}{u} du - \displaystyle\int_1^x \frac{\log(u)\left\{u\right\}}{u^2} du$

$=\log(x)-\frac{\left\{x\right\}\log(x)}{x}-\log(x) + \log(1)+ \displaystyle\int_1^x \frac{\left\{u\right\}}{u^2} du+ \frac{1}{2}\log^2(x)-\frac{1}{2}\log^2(1)- \displaystyle\int_1^x \frac{\log(u)\left\{u\right\}}{u^2} du$

$=-\frac{\left\{x\right\}\log(x)}{x}+\displaystyle\int_1^x \frac{\left\{u\right\}}{u^2} du+ \frac{1}{2}\log^2(x)- \displaystyle\int_1^x \frac{\log(u)\left\{u\right\}}{u^2} du$.

Then, $-\frac{\left\{x\right\}\log(x)}{x}=O\left(\frac{\log(x)}{x}\right)$,

$\displaystyle\int_1^x \frac{\left\{u\right\}}{u^2} du=O\left(\frac{1}{x}\right)$,

and $ - \displaystyle\int_1^x \frac{\log(u)\left\{u\right\}}{u^2} du= O\left(\frac{\log(x)}{x}\right)$

In the end, I obtain that the sum is equal to $\frac{1}{2}\log^2(x) + O\left(\frac{\log(x)}{x}\right)$. Then I have tried to plot my result to check it, but it doesn't seem a correct extimation. I would like to know if I am doing something wrong or if there is another approach to the problem. Thank you

Edit: I think that the last two errors are wrong. $\displaystyle\int_1^x \frac{\left\{u\right\}}{u^2} du \leq \displaystyle\int_1^x \frac{{1}}{u^2}=-\frac{1}{x}+1=O\left({1}\right)$,

and $ - \displaystyle\int_1^x \frac{\log(u)\left\{u\right\}}{u^2} du\leq \displaystyle\int_1^x \frac{\log(u)}{u^2} du= 1- \frac{1}{x}-\frac{\log(x)}{x} = O\left(1\right)$

So I would write that the sum is equal to $\frac{1}{2}\log^2(x) + O\left(1\right)$. Thanks for your patience

Answer 1

Let me tell you something that makes nearly all such problems (at a basic level) straightforward without having to apply partial summation (Abel summation formula) directly.

Theorem. If $a_n$ and $b_n$ are sequences of positive numbers and $a_n \sim b_n$ with $\sum_{n \geq 1} b_n = \infty$, then $\sum_{n \leq N} a_n \sim \sum_{n \leq N} b_n$ as $N \to \infty$.

Proof. Left to the reader. It's a nice exercise.

Clearly it's not important that the sequence starts at $n = 1$ in this theorem. Any starting point is fine.

Let's see how the theorem lets us replace sums with integrals to make asymptotic estimates under mild (and practical) conditions.

When $f(t)$ is a continuous positive function that is monotonic for large $t$ and $f(n) \sim f(n+1)$ as $n \to \infty$, we have $$ f(n) \sim \int_n^{n+1} f(t)\,dt $$ by making upper and lower bound estimates on the integral using $f(n)$ and $f(n+1)$. Since $$ \sum_{n=1}^N \int_n^{n+1} f(t)\,dt = \int_1^{N+1} f(t)\,dt, $$ if $\int_1^\infty f(t)\,dt = \infty$ we can use the theorem with $a_n = f(n)$ and $b_n= \int_n^{n+1} f(t)\,dt$ to obtain $$ \sum_{n\leq N} f(n) \sim \int_1^{N+1}f(t)\,dt. $$ It's unimportant that the summing and integrating starts at $1$. It could start at $2$, $3$, and so on.

Example. Let $f(t) = (\log t)/t$ for $t \geq 2$. This is continuous, positive, and monotonic for $t \geq 3$, and $$ \int_1^\infty \frac{\log t}{t}\,dt = \left.\frac{1}{2}(\log t)^2\right|_1^\infty = \infty, $$ so as $N \to \infty$ in $\mathbf Z$, $$ \sum_{n \leq N} \frac{\log n}{n} \sim \int_1^{N+1} \frac{\log t}{t}\,dt = \left.\frac{1}{2}(\log t)^2\right|_1^{N+1} = \frac{1}{2}(\log(N+1))^2. $$ Since $\log(N+1) \sim \log N$, the above estimate implies $$ \sum_{n \leq N} \frac{\log n}{n} \sim \frac{1}{2}(\log N)^2, $$ and since $\log x \sim \log \lfloor x\rfloor$, we also obtain $$ \sum_{n \leq x} \frac{\log n}{n} \sim \frac{1}{2}(\log x)^2 $$ as $x \to \infty$ in $\mathbf R$.

Remark. This way of using the theorem above makes it really easy to find asymptotic estimates for partial sums in simple situations without having to think too hard. But it does not provide lower-order terms in the estimates. For that, you'll want to use partial summation.

Exercise. Show by the same method that for each $\alpha > 0$, $$ \sum_{n \leq x} \frac{(\log n)^\alpha}{n} \sim \frac{1}{\alpha+1}(\log x)^{\alpha+1} $$ as $x \to \infty$. In fact, the work should apply for $\alpha > -1$, not just $\alpha > 0$. (It would not make sense when $\alpha = -1$.)

Answer 2

Using the fact that $$ 1+\frac12+\dots+\frac1n=\ln n+\gamma+o(1) $$ where $\gamma$ is the Euler-Mascheroni constant, we can write that $$ f(m)=\sum_{n=1}^m \frac{\ln n}n \asymp \sum_{n=1}^m \frac1n\left[1+\frac12+...+\frac1n-\gamma\right] =-\gamma \sum_{n=1}^m \frac1n +\frac12\left[\left(\sum_{n=1}^m\frac1n\right)^2-\sum_{n=1}^m\frac1{n^2}\right] =\frac{1+o(1)}2\left(\sum_{n=1}^m\frac1n\right)^2 $$ hence $f(m)\sim \frac12 (\ln m)^2$.

Answer 3

Use Abel-Plana formula, Eq.(2.10.2) $$\sum_{n=a}^x f(n)=\int_a^x f(x)dx+\frac{1}{2}f(a)+\frac{1}{2}f(x)-2\int_0^\infty \frac{\Im f(a+it)}{e^{2\pi t}-1}dt+\hat{R}$$

where $\hat{R}$ is the remainder representing the last two terms in the second line of Eq.(2.10.2).

Let $$f(t)=\frac{\ln t}{t}$$

then the remainder $\hat{R}\sim o\left(\frac{\ln x}{x}\right)$ when $x\rightarrow\infty$ for our choice of $f(t)$.

Set $a=1$, we get

$$f(1+it)=\frac{\ln(1+it)}{1+it}=\frac{\ln\sqrt{1+t^2}+i\arctan t}{1+it}$$

The imaginary part:

$$\Im f(1+it)=\arctan t-\frac{t}2\cdot\ln(1+t^2)$$

It is easy to show the integral converges, hence it equals some constant $C$.

$$\int_0^\infty \frac{\Im f(a+it)}{e^{2\pi t}-1}dt=\int_0^\infty \frac{\arctan t}{e^{2\pi t}-1}dt-\int_0^\infty \frac{t\cdot\ln(1+t^2)}{2(e^{2\pi t}-1)}dt=C$$

So for large $x$, we get

$$\boxed{\sum_{n=1}^x \frac{\ln n}{n}\sim\int_1^x \frac{\ln t}{t}dt+C+\frac{\ln x}{2x}=\frac{1}2\ln^2(x)+C+\frac{\ln x}{2x}}$$

If we consider the leading order in the remainder $\hat{R}$, we can get

$$\boxed{\sum_{n=1}^x \frac{\ln n}{n}\sim\frac{1}2\ln^2(x)+C+\frac{1}2\cdot\frac{\ln x}{x}-\frac{1}{12}\cdot\frac{\ln x}{x^2}+\frac{1}{12}\cdot\frac{1}{x^2}+\dots}$$

Answer 4

To give more than the asymptotics

Since $$\int_1^x \frac {\log(n)}n \,dn=\frac{1}{2}\log ^2(x)$$ using Euler-MacLaurin summation formula, we have $$\displaystyle\sum_{n=1}^x \frac{\log(n)}{n}=C+\frac{\log ^2(x)}{2}+\frac{\log (x)}{2 x}-\frac{\log (x)-1}{12 x^2}+\frac{6 \log (x)-11}{720 x^4}-\frac{60 \log (x)-137}{15120 x^6}+O\left(\frac{1}{x^8}\right)$$ On the other hand, we have $$\displaystyle\sum_{n=1}^x \frac{\log(n)}{n}=\gamma _1-\gamma _1(x+1)$$ where appear the generalized Stieltjes constant.

Using, as a reference $x=10^3$ gives as a rationalized number $C=-\frac{15}{206}$ which can be neglected.