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.

  • 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 $0.386294$.

  • 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.

Solutions Collecting From Web of "What is $\lim_{n \to \infty} \sum_{x=0}^{n-1} \frac{n-x}{n+x}$?"

$$\sum_{x=0}^{n-1} \frac{n-x}{n+x} $$
$$=\sum_{x=0}^{n-1} \frac{-(n+x)+2n}{n+x}$$
$$=-\sum_{x=0}^{n-1} \frac{n+x}{n+x} +2\sum_{x=0}^{n-1} \frac{n}{n+x}$$
$$=-n+2n\sum_{x=0}^{n-1} \frac{1}{n+x}$$
$$=-n+2n\sum_{x=n}^{2n-1} \frac{1}{x}$$
$$=-n +2n(H_{2n-1}-H_{n-1})$$

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 $\psi(k+1) – \psi(k) =\frac{1}{k}$.

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.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\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}