What is $\lim_{n \to \infty} \sum_{x=0}^{n-1} \frac{n-x}{n+x}$? These are two little questions that came to mind while I was looking at this problem.


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*What is $\displaystyle \lim_{n \to \infty} \sum_{x=0}^{n-1} \frac{n-x}{n+x}$?


I am fairly certain that the answer is $\infty$ because as $n$ gets closer to $\infty$ there are more terms that are very close to $1$ (if $n = 1,000,000$ then all the terms until $x = 5026$ are greater than or equal to $0.99$, and if $n = 1,000,000,000$ then you have to get to $x = 5025126$ for the terms to drop below $0.99$), but I don't know how to prove it.
I also checked the partial differences (i.e. between $n = 1$ and $n = 2$, between $n = 2$ and $n = 3$, and so forth) and noticed that they all tend to some number around $0.386294$.


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*Is there a name for this number, and what's its significance? WolframAlpha seems to suggest it has something to do with the Digamma function but I'm not sure what it's all about.

 A: $$\sum_{x=0}^{n-1} \frac{n-x}{n+x} $$
$$=\sum_{x=0}^{n-1} \frac{-(n+x)+2n}{n+x}$$
$$=-\sum_{x=0}^{n-1} \frac{n+x}{n+x} +2\sum_{x=0}^{n-1} \frac{n}{n+x}$$
$$=-n+2n\sum_{x=0}^{n-1} \frac{1}{n+x}$$
$$=-n+2n\sum_{x=n}^{2n-1} \frac{1}{x}$$
$$=-n +2n(H_{2n-1}-H_{n-1})$$
Where $H_k$ is the $k$th harmonic number. The harmonic numbers are indeed connected to the digamma function $\psi$ by
$$H_k = \gamma + \psi(k+1),$$
where $\gamma$ is the Euler-Mascheroni constant. Now, using properties of the digamma function (look: http://mathworld.wolfram.com/DigammaFunction.html or http://en.wikipedia.org/wiki/Digamma_function), we have
$$H_{2n-1}-H_{n-1} = \psi(2n) - \psi(n) = \frac{1}{2}\psi(n+\frac{1}{2}) - \frac{1}{2}\psi(n)+\ln 2 \geq \ln 2$$
So we see that the original sum tends to infinity as $n\to \infty.$
The difference between the sum for $n$ and $n-1$ is
$$-n +2n(H_{2n-1}-H_{n-1}) + (n-1)  - 2(n-1)(H_{2(n-1)-1}-H_{(n-1)-1})$$
$$= 2n(H_{2n-1}-H_{n-1}-H_{2n-3}+H_{n-2}) + 2(H_{2(n-1)-1}-H_{(n-1)-1}) -1 $$
$$= 2n(\frac{1}{2n-1}+\frac{1}{2n-2}-\frac{1}{n-1}) +  2(H_{2(n-1)-1}-H_{(n-1)-1}) -1$$
$$=-\frac{2n}{4 n^2-6 n+2} +  2(H_{2(n-1)-1}-H_{(n-1)-1}) -1$$
Now as $n\to \infty$, the first part goes to $0$ and the second to $2\ln 2 -1 \approx 0.386294$. The second part was already calculated above (for n) and $\frac{1}{2}\psi(n+\frac{1}{2}) - \frac{1}{2}\psi(n)$ tends to zero, since $\psi$ is monotonic in $\mathbb{R}_+$ and $\psi(k+1) - \psi(k) =\frac{1}{k}$.
A: You can use the integral approach as

$$ \sum_{k=0}^{n-1}\frac{n-t}{n+t} \sim \int_{0}^{n-1}\frac{n-t}{n+t} dt. $$ 

See the main result for integral test.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}
\color{#66f}{\large\lim_{n \to \infty}\sum_{x = 0}^{n - 1}{n - x \over n + x}}
&=\lim_{n \to \infty}\bracks{\sum_{x = 0}^{n - 1}{2n \over n + x} - n} 
=\lim_{n \to \infty}\bracks{2n\sum_{x = 0}^{n - 1}{1/n \over 1 + x/n} - n}
\\[3mm]&=\lim_{n \to \infty}\bracks{2n\int_{0}^{1 - 1/n}{\dd x \over 1 + x} - n}
=\lim_{n \to \infty}\bracks{2n\ln\pars{2 - {1 \over n}} - n}
=\color{#66f}{\large\infty}
\end{align}
