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.
  • $\begingroup$ The limit goes to infinity unless you are missing something. $\endgroup$ Feb 18 '14 at 21:10
  • $\begingroup$ By Contradiction! I think you can formulate your argument and come up with a proof by contradiction, although it still needs some thoughts. $\endgroup$
    – Alt
    Feb 18 '14 at 21:11

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

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

  • $\begingroup$ Wow, very thorough. Thanks a lot! I had a hunch that the constant would be related to $\ln 2$ but I wasn't exactly sure how. $\endgroup$
    – 2012ssohn
    Feb 19 '14 at 0:53
  • $\begingroup$ @ploosu2 What is the answer ?. I was lost. $\endgroup$ Jul 25 '14 at 4:23

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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.