Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I show that:

There exists a bijective $f: \mathbb{N} \to \mathbb{N}$ such that: $$\sum_{n=1}^{\infty} (-1)^{f(n)}\ln\frac{f(n)+1}{f(n)}=\ln 2010$$

I have really no idea where to begin... Taylor series doesn't seem related. I thought about the alternating series test which approximates the sum but not sure how to use it.

Any help will be appreciated.

share|cite|improve this question
Have you tried taking the exponential of both sides? – Qiaochu Yuan Jan 7 '11 at 14:29
Now that you mention it, it seems obvious. Taking the exponential gives $\sum_{n=1}^{\infty} (1+\frac{1}{f(n)})^{(-1)^{f(n)}}=2010$. I'll think about it a little, nothing pops to mind yet. – daniel.jackson Jan 7 '11 at 14:53
Not quite. The sum turns into a product. – Qiaochu Yuan Jan 7 '11 at 15:49
up vote 10 down vote accepted

Suppose first you take $f(n)=2n$ then $$\sum_{n=1}^{\infty} (-1)^{2n}\ln\frac{2n+1}{2n}= \sum_{n=1}^{\infty} \ln(2n+1)-\ln(2n) = \sum_{n=1}^{\infty} \frac {1}{2n+\delta_n} \geq \sum_{n=1}^{\infty} \frac {1}{2n+1} = \infty$$ where $0<\delta_n<1$ - this is Lagrange's theorem $\frac{f(x) - f(y)}{(x)-y)} = f' (c)$ for some $c\in (x, y)$, and here $x=2n+1,\; y=2n,\; f(t)=\ln(t)$ so $f'(t)=\frac{1}{t}$.

The same works for $f(n)=2n+1$ but with $-\infty$. Now construct another function $f$ as follows : choose enough even numbers until the first time your partial sum is above $\ln 2100$, then choose enough odd numbers until the first time your partial sum is below $\ln 2100$. you can do this since the two sequences I described above converge to $\pm \infty$. you can continue like this to build a bijective map $f:\mathbb{N}\rightarrow \mathbb{N}$.

Since the $n-th$ term in the sum converges to zero, show that the sequence converges to $\ln 2100$.

share|cite|improve this answer
@Promotheus: Very nice! Thank you for the argument. – Jonas Teuwen Jan 7 '11 at 14:45
Thanks. I'm not sure I'm familiar with Lagrange's theorem, and there seem to be a few, which one did you mean? Also, the question explicitly states "there's no need to look for $f(n)$" ;) but I'll try to understand your argument nonetheless. – daniel.jackson Jan 7 '11 at 14:49
This is essentially a proof of the Riemann rearrangement theorem, which the question is a special case of. – Chris Eagle Jan 7 '11 at 15:12
ofcourse, I completely forgot about Riemann rearrangement theorem! I can take $f(n)=n$ and show that $\sum (-1)^n \ln (1+\frac{1}{n})$ converges conditionally and then use the theorem. – daniel.jackson Jan 7 '11 at 15:21
@Prometheus: I think most people call your Lagrange's theorem the mean value theorem. – Chris Eagle Jan 7 '11 at 15:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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