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

Question

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.

Answer 1

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

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 $\psi(k+1) - \psi(k) =\frac{1}{k}$.

Answer 2

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.

Answer 3

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