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I am interested in the limit

$\lim_{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{n} \frac{k}{\ln(k+1)}$.

While I suspect the limit to be 0, I cannot prove it rigorously. By intuition we have $\sum_{k=1}^{n} k \sim n^2$, so by making each term a bit smaller, we can achieve the limit to be zero. However, due to the nature of logarithms, it is difficult to make an estimation. For example, we have $\ln(k+1) \geq \frac{k}{k+1}, \, \forall k \in \mathbb{N}$, but then $\sum_{k=1}^{n} \frac{k}{\ln(k+1)} \leq \sum_{k=1}^{n} k \frac{k+1}{k} = \sum_{k=1}^{n} (k+1) \sim n^2$, so we are back at square one. Does the limit exist, and if so, how do I prove the statement?

user21820
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Tom Lucas
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  • Using integral: $$\underset{n\to \infty }{\text{lim}}\frac{\int_1^n \frac{k}{\log (k+1)} , dk}{n^2}=\underset{n\to \infty }{\text{lim}}\frac{\text{Ei}(2 \log (1+n))+\text{li}(2)-\text{li}(4)-\text{li}(1+n)}{n^2}=0$$ – Mariusz Iwaniuk Jul 01 '24 at 21:30
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    $$\sum_{k=1}^n \frac k{\ln(k+1)}\sim\frac{n^2}{2\ln n}\sum_{k=0}^\infty\frac{k!}{(2\ln n)^k}=\frac{n^2}{2\ln n}+\frac{n^2}{4\ln^2 n}+O\left(\frac{n^2}{\ln^3n}\right);,n\to\infty$$ – Svyatoslav Jul 02 '24 at 01:03
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    Two more duplicates: https://math.stackexchange.com/a/3859004 and https://math.stackexchange.com/a/1552743 – Anne Bauval Jul 02 '24 at 09:58

3 Answers3

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Let $f(x)=\dfrac{x}{\ln(1+x)}$.

We have that $$f'(x)=\dfrac{\ln(1+x)-\frac{x}{1+x}}{\ln^2(1+x)}\geq 0$$

so that $f$ is increasing.

Thus, for all $n\geq 1$ we have $$0\leq \frac{1}{n^2} \sum_{k=1}^{n} \frac{k}{\ln(k+1)}\leq \frac{1}{n^2} \sum_{k=1}^{n} \frac{n}{\ln(n+1)}=\dfrac{1}{\ln(n+1)}\to 0$$

By the squeeze theorem we find $$\lim_{n\to\infty} \frac{1}{n^2} \sum_{k=1}^{n} \frac{k}{\ln(k+1)}=0,$$

as expected.

Julio Puerta
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We have $${1\over n^2}\sum_{k=1}^n{k\over \ln(k+1)}\le {1\over n}\sum_{k=1}^n{1\over \ln(k+1)}$$ Now we apply the well known fact: if $a_n\to 0$ then ${1\over n}\sum_{k=1}^na_k\to 0,$ with $a_n={1\over\ln(n+1)}.$

Ryszard Szwarc
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$$I_n=\int_{1}^{n} \frac k {\log(k+1)}\,dk<\sum_{k=1}^n \frac k {\log(k+1)} <\int_{1}^{n+1} \frac k {\log(k+1)}\,dk=I_{n+1}$$

$$I=\int \frac k {\log(k+1)}\,dk= \int\frac {k+1} {\log(k+1)}\,dk-\int\frac {1} {\log(k+1)}\,dk$$ $$I=\text{Ei}(2 \log (k+1))-\text{li}(k+1)$$ where $\text{Ei}(.)$ is the exponential integral function and $\text{li}(.)$ the logarithm integral function.

Using asymptotics to first order $$\frac{I_1}{n^2}=\frac{I_2}{n^2}=\frac{4 \log ^3(n)+2 \log ^2(n)+2 \log(n)+3}{8 \log^4(n)}+O\left(\frac{1}{n\log(n)}\right)\sim \frac 1{2\log(n)}$$

For $n=1000$, the above formula gives $0.0785448$ while the exact sumaation would give $0.0786658$ (relative error of $0.15$%).

Claude Leibovici
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