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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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    Not quite. The sum turns into a product.2011-01-07

1 Answers 1

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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 $\ln 2100$.

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    @Prometheus: I think most people call your Lagrange's theorem the [mean value theorem](http://en.wikipedia.org/wiki/Mean_value_theorem).2011-01-07