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

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.

• How did you conclude that $\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\frac{\frac{1}{n}}{\frac{k}{n} (2-\frac{k}{n}+\frac{1}{n})}=\int_{0}^{1} \frac{1}{x(2-x)}dx$? – Zacky Mar 4 '19 at 8:27
• – Septimiu Cristian Mar 4 '19 at 8:29
• You seem to have ignored the red part:$$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}+\color{red}{\frac{1}{n}}\right)} \neq \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}\right)}=\int_0^1 \frac{dx}{x(2-x)}$$ – Zacky Mar 4 '19 at 8:30
• $$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}+\color{red}{\frac{1}{n}}\right)}$$ That is another problem I stumbled upon. I suppose it dissapears as n increases towards $\infty$ or it just becomes $\frac{k-1}{n}$ but I am not sure. – Septimiu Cristian Mar 4 '19 at 8:32

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}

• Could you help me with the late edit I added? – Septimiu Cristian Mar 4 '19 at 8:50
• @SeptimiuCristian help yourself with the Euler-Mascheroni constant - the definition is very similar to your expression. en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant – orion Mar 4 '19 at 8:51
• @SeptimiuCristian Sure. See my edit. – Robert Z Mar 4 '19 at 9:11

$$2n-k+1 \leq 2n$$ so $$\sum\limits_{k=1}^{n} \frac n {k(2n-k+1)} \geq \sum\limits_{k=1}^{n} \frac n {2kn}= \sum\limits_{k=1}^{n} \frac 1 {2k} \to \infty$$.
• @SeptimiuCristian If $x_n \geq y_n$ for all $n$ and $y_n \to \infty$ then $x_n \to \infty$. – Kavi Rama Murthy Mar 4 '19 at 8:44
• @SeptimiuCristian $y_n \to \infty$ implies that $y_n >0$ for $n$ sufficiently large and hence $x _n \geq y_n >0$ for $n$ sufficiently large so we are not concerned with negative numbers here. Actually all the numbers in the question are positive. – Kavi Rama Murthy Mar 4 '19 at 8:47