# Show that there's a bijective $f: \mathbb{N} \to \mathbb{N}$ such that $\sum_{n=1}^{\infty} (-1)^{f(n)}\ln\frac{f(n)+1}{f(n)}=\ln 2010$

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.

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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

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 $\ln 2100$.
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
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