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Discounted Cumulative Gain (in the simpler, binary relevance version) is defined as

$$ DCG_{\text{binary}} := \sum_{i=1}^M \frac{r_i}{\log(i + 1)} $$ with $r_i\in\{0,1\}$. (Since $2^r-1=r$ for both 0 and 1)

It is commonly normalized with the maximum achievable score called $IDCG$, obtained by sorting the $r_1$ in descending order, which for binary effectively is: $r_i = \begin{cases}1 & i \leq n\\0 & i > n\end{cases}$. Then $$ IDCG_{\text{binary}} := \sum_{i=1}^{n} \frac{1}{\log(i + 1)} = \sum_{i=2}^{n+1} \frac{1}{\log(i)} $$ Rather than computing this with a loop, I am therefore looking for a closed form solution of $\sum_{i=2}^n 1/\log(i)$. I could not figure one out myself.

By exponentiation I got this alternate form: $$ \exp^{IDCG_{\text{binary}}} = \exp^{\sum_{i=2}^{n+1} \frac{1}{\log(i)}} \\ = \prod_{i=2}^{n+1} \exp\frac{1}{\log(i)} $$ but that isn't much nicer.

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    Your transformation of sum into product is wrong $$ \exp(\sum_i 1/\log(i)) = \prod_i \exp(1/\log(i)) \neq \prod_i1/\log(i) $$ – 2017-02-10
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    Thanks. Just ignore that part then. It did not work anyway. – 2017-02-10
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    I'm afraid that the sum $\sum_i 1/\log(i)$ has no closed form. One way to approximate it would be with the integral $\int_2^{n+1} 1/\log(x) dx$ however this integral doesn't a closed form either. It is actually defined to be the logarithm-integral $Li(n+1)$. – 2017-02-10

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