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How to prove:

$$ \sum_{n=0}^{\infty}\left[\,\sum_{k=1}^{a}\frac{1}{a\,n+k}-\sum_{k=1}^{b}\frac{1}{b\,n+k}\,\right]=\log\left(\frac{a}{b}\right)\qquad\colon\,a\,,b\in\mathbb{N}^{+}\tag{1}$$

Obviously, we cannot split the sum: $$ \left(\sum_{n=0}^{\infty}\,\sum_{k=1}^{a}\frac{1}{a\,n+k}\right)-\left(\sum_{n=0}^{\infty}\,\sum_{k=1}^{b}\frac{1}{b\,n+k}\right) = \left(\sum_{n=1}^{\infty}\frac1n\right)-\left(\sum_{n=1}^{\infty}\frac1n\right)=\infty-\infty $$ Any ideas? Thanks in advance.

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For starters, let us give an integral representation to the general term of the outer sum:

$$ \sum_{k=1}^{a}\frac{1}{an+k}-\sum_{k=1}^{b}\frac{1}{bn+k} = \int_{0}^{1}\left(x^{an}\sum_{k=1}^{a}x^{k-1}-x^{bn}\sum_{k=1}^{b}x^{k-1}\right)\,dx \tag{1}$$ turning the whole sum into: $$ \lim_{N\to +\infty}\sum_{n= 0}^{N}\int_{0}^{1}\left[\left(x^{an}-x^{bn}\right)-\left(x^{a(n+1)}-x^{b(n+1)}\right)\right]\,\frac{dx}{1-x} \tag{2}$$ If now we exploit $$ \int_{0}^{1}\frac{x^A-x^B}{1-x}\,dx = H_B-H_A \tag{3}$$ the whole sum turns into $$ \lim_{N\to +\infty} \left(H_{aN}-H_{bN}\right) = \log\frac{a}{b} \tag{4}$$ due to $H_n=\log(n)+\gamma+O\left(\frac{1}{n}\right).$


It is interesting to point out that we are not allowed to exchange $\sum_{n\geq 0}$ and $\int_{0}^{1}$ in $(2)$, since the hypothesis of the dominated convergence theorem are not met.

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  • $\begingroup$ I was just typing the same approach (+1). $\endgroup$ – Zaid Alyafeai Feb 28 '17 at 22:36
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    $\begingroup$ You don't have to get an integral out to do this: you can see just by writing out the sums carefully that the partial sum is $H_{aN}-H_{bN}$. $\endgroup$ – Chappers Feb 28 '17 at 22:40
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    $\begingroup$ Your answering speed is away beyond expectations (^_^)! Thanks Jack. (+1) $\endgroup$ – Hazem Orabi Feb 28 '17 at 22:43

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