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$$\sum_{n=1}^{2^k - 1}\frac{1}{n} = \sum_{n=1}^{2-1}\frac{1}{n} + \sum_{n=2}^{2^2 - 1}\frac{1}{n} + \cdots + \sum_{n = 2^{k-1}}^{2^k - 1}\frac{1}{n} = \boxed{\sum_{j = 0}^{k-1}\sum_{n = 2^j}^{2^{j+1} - 1}\frac{1}{n}}$$

I haven't seen two summation symbols beside each other before, I would really appreciate it if someone could tell me what the last part that I have highlighted means.

I know the first part (before the first equality) is the summation of the harmonic series from $n=1$ to $(2^k)-1$. The middle is the partial sums from $n=1$ to $n=(2^k)-1$. But after the final equality I'm not sure?

I will at least say what I think it means, the summation of the partial sums of the harmonic series from $n=2^j$ to $(2^j)-1$, and to take those sums from $j=0$ to $k-1$.

Any help would be greatly appreciated, thanks in advance

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  • $\begingroup$ You're right. It means what you think it means. $\endgroup$ – Ken Wei Oct 31 '17 at 4:20
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You're exactly right. It might help to picture the implied parentheses: $$\sum_{j=0}^{k-1} \sum_{n=2^j}^{2^{j+1} - 1} \frac{1}{n} = \sum_{j=0}^{k-1} \left (\sum_{n=2^j}^{2^{j+1} - 1} \frac{1}{n}\right)$$ They are pretty much always omitted for convenience, the same way multiple integrals don't have nested parentheses.

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