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I am trying to proof: $$\lim_{n\to \infty}\sum_{k=0}^{n-1}\frac{1}{n}\left(\ln\left(\frac{m}{n}+\frac{k}{n}\right)\right)=\int_{0}^{1}\ln(x)dx \hskip50pt (1)$$

where $m\in\mathbb{R}$ and $m$ positive.

I got inspiration in this post: Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ and the answer Prove that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k = 1}^{n}{f(\frac{k}{n}) }$ $=\int_0^1 f(x)dx.$ but I am not sure about how to deal with the $\frac{m}{n}$ term since my partition $\{0 +\frac{m}{n},\frac{1}{n}+\frac{m}{n},\frac{2}{n}+\frac{m}{n},\dots,\frac{(n-1)}{n},1+\frac{m}{n}\}$ isn`t from $0$ to $1$.

Maybe (1) is false? Am I missing something?

Any hint is appreciated.

Yiorgos S. Smyrlis
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mathstudent007
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1 Answers1

1

Since $\ln x$ is strictly increasing, then $$ \frac{1}{n}\ln\left(\frac{k}{n}\right)<\int_{k/n}^{(k+1)/n}\ln x\,dx <\frac{1}{n}\ln\left(\frac{k+1}{n}\right) $$ and hence $$ \frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k+1}{n}\right) < \frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\int_{k/n}^{(k+1)/n}\ln x\,dx < \frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k}{n}\right) $$ But $$ \frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k}{n}\right) =\frac{1}{n}\ln\left(\frac{m+k}{k}\right)< \frac{1}{n}\ln\left(1+\frac{m}{k}\right)<\frac{m}{kn}, $$ while $$ \frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\frac{1}{n}\ln\left(\frac{k+1}{n}\right)=\frac{1}{n}\ln\left(\frac{m+k}{k+1}\right)= \frac{1}{n}\ln\left(1+\frac{m-1}{k+1}\right) $$ If $m\ge 1$, then $\ln\left(1+\frac{m-1}{k+1}\right)\ge 0$. If $m\in(0,1)$, then $$ \ln\left(1+\frac{m-1}{k+1}\right)=\frac{1}{k+1}\ln\left(1+\frac{m-1}{k+1}\right)^{k+1}\ge \frac{1}{k+1}\ln\left(1+(k+1)\frac{m-1}{k+1}\right)=\frac{\ln m}{k+1}. $$ Thus $$ \frac{1}{(k+1)n}\min\{0,\ln m\}<\frac{1}{n}\ln\left(\frac{m+k}{n}\right)-\int_{k/n}^{(k+1)/n}\ln x\,dx<\frac{m}{kn} $$ and therefore $$ \frac{\min\{0,\ln m\}}{n}(\ln(n+1)-\ln 2) < \frac{\min\{0,\ln m\}}{n}\sum_{k=1}^{n-1}\frac{1}{(k+1)}<\sum_{k=1}^{n-1}\frac{1}{n}\ln\left(\frac{m}{n}+\frac{k}{n}\right) -\int_{1/n}^1\ln x\,dx<\sum_{k=1}^{n-1}\frac{m}{kn} =\frac{m}{n}\sum_{k=1}^{n-1}\frac{1}{k} <\frac{m}{n}(1+\ln n) $$ since $1+\ln n>\sum_{k=1}^{n-1}\frac{1}{k}$.

Yiorgos S. Smyrlis
  • 83,933