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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.

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    $\begingroup$ How did you conclude that $\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\frac{\frac{1}{n}}{\frac{k}{n} (2-\frac{k}{n}+\frac{1}{n})}=\int_{0}^{1} \frac{1}{x(2-x)}dx$? $\endgroup$ – Zacky Mar 4 '19 at 8:27
  • $\begingroup$ @Zacky mathworld.wolfram.com/RiemannSum.html $\endgroup$ – Septimiu Cristian Mar 4 '19 at 8:29
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    $\begingroup$ You seem to have ignored the red part:$$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}+\color{red}{\frac{1}{n}}\right)} \neq \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}\right)}=\int_0^1 \frac{dx}{x(2-x)}$$ $\endgroup$ – Zacky Mar 4 '19 at 8:30
  • $\begingroup$ $$\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\frac{1}{\frac{k}{n} \left(2-\frac{k}{n}+\color{red}{\frac{1}{n}}\right)}$$ 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. $\endgroup$ – Septimiu Cristian Mar 4 '19 at 8:32
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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: $$\begin{align}\sum_{k=1}^{n}\frac{n}{k(2n-k+1)}-\frac{\ln(n)}{2}&=\frac{nH_{2n}}{2n+1}-\frac{\ln(n)}{2}\\ &=\frac{n(\ln(2n)+\gamma+o(1))}{2n+1}-\frac{\ln(n)}{2}\\ &=\frac{n(\ln(2)+\ln(n)+\gamma+o(1))}{2n+1}-\frac{\ln(n)}{2}\to \frac{\ln(2)+\gamma}{2}.\end{align}$$

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  • $\begingroup$ Could you help me with the late edit I added? $\endgroup$ – Septimiu Cristian Mar 4 '19 at 8:50
  • $\begingroup$ @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 $\endgroup$ – orion Mar 4 '19 at 8:51
  • $\begingroup$ @SeptimiuCristian Sure. See my edit. $\endgroup$ – Robert Z Mar 4 '19 at 9:11
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Please note: OP has added another part to the question much after two answers appeared.

$2n-k+1 \leq 2n$ so $\sum\limits_{k=1}^{n} \frac n {k(2n-k+1)} \geq \sum\limits_{k=1}^{n} \frac n {2kn}= \sum\limits_{k=1}^{n} \frac 1 {2k} \to \infty$.

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  • $\begingroup$ Isn't the harmonic series diverging? $\endgroup$ – Septimiu Cristian Mar 4 '19 at 8:34
  • $\begingroup$ Yes, that is exactly what I am using. $\endgroup$ – Kavi Rama Murthy Mar 4 '19 at 8:35
  • $\begingroup$ I am not sure that from comparing a sum with a diverging series you obtain that the first sum is diverging too. You can construct a converging curve over the graph of a diverging series and by this prove it's not sufficient to make such comparison to conclude the first function diverges too. $\endgroup$ – Septimiu Cristian Mar 4 '19 at 8:42
  • $\begingroup$ @SeptimiuCristian If $x_n \geq y_n$ for all $n$ and $y_n \to \infty$ then $x_n \to \infty$. $\endgroup$ – Kavi Rama Murthy Mar 4 '19 at 8:44
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Mar 4 '19 at 8:47

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