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

Question

I have to find $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{n}{k(2n-k+1)}$. This limit is equal to $\lim_{n\rightarrow \infty} \sum_{k=1}^{n}(\frac{\frac{1}{n}}{\frac{k}{n} (2-\frac{k}{n}+\frac{1}{n})})$=$\int_{0}^{1} \frac{1}{x(2-x)}dx$. After decomposing the last rational fraction, I have problems integrating $\int_{0}^{1}\frac{1}{x}$ because I don't know what is $\ln(0)$. Any help, please?

Late(r) edit: What is then $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{n}{k(2n-k+1)}-\frac{1}{2}\ln(n)$. I assumed that I will be able to handle it after finding out the previous limit.

Answer 1

Note that $$\frac{n}{k(2n-k+1)}=\frac{n}{2n+1}\left(\frac{1}{k}+\frac{1}{2n+1-k}\right)$$ Therefore, as $n$ goes to infinity $$\sum_{k=1}^{n}\frac{n}{k(2n-k+1)}=\underbrace{\frac{n}{2n+1}}_{\to 1/2}\cdot\underbrace{\sum_{k=1}^{2n}\frac{1}{k}}_{\to +\infty}\to +\infty$$ because the harmonic sequence $H_{m}=\sum_{k=1}^{m}\frac{1}{k}$ is divergent.

As regards the second limit, use $H_{m}=\ln(m)+\gamma+o(1)$ where gamma is the Euler-Mascheroni constant: $$\begin{align}\sum_{k=1}^{n}\frac{n}{k(2n-k+1)}-\frac{\ln(n)}{2}&=\frac{nH_{2n}}{2n+1}-\frac{\ln(n)}{2}\\ &=\frac{n(\ln(2n)+\gamma+o(1))}{2n+1}-\frac{\ln(n)}{2}\\ &=\frac{n(\ln(2)+\ln(n)+\gamma+o(1))}{2n+1}-\frac{\ln(n)}{2}\to \frac{\ln(2)+\gamma}{2}.\end{align}$$

Answer 2

Please note: OP has added another part to the question much after two answers appeared. $2n-k+1 \leq 2n$ so $\sum\limits_{k=1}^{n} \frac n {k(2n-k+1)} \geq \sum\limits_{k=1}^{n} \frac n {2kn}= \sum\limits_{k=1}^{n} \frac 1 {2k} \to \infty$.