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What's it equal to: $$\lim_{n\rightarrow \infty }\sum_{1\leqslant k\leqslant n}\frac{1}{k\ln (n+k)}$$

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up vote 4 down vote accepted

By partial summation, $$\sum_{k=1}^{n}\frac{1}{k\log(n+k)}=\frac{H_n}{\log(2n)}-\sum_{k=1}^{n-1}H_k\left(\frac{1}{\log(n+k+1)}-\frac{1}{\log(n+k)}\right),$$ but the last term is $O\left(\frac{1}{\log n}\right)$ while the limit of the first term is just $1$.

As an alternative, consider that $$\frac{H_n}{\log(2n)}\leq\sum_{k=1}^{n}\frac{1}{k\log(n+k)}\leq \frac{H_n}{\log n}$$ is trivial, while the Cauchy-Schwarz inequality gives: $$\sum_{k=1}^{n}\frac{1}{k\log(n+k)}\geq\frac{H_n^2}{\sum_{k=1}^n \frac{\log(n+k)}{k}}=\frac{H_n^2}{H_n \log n+\sum_{k=1}^n\frac{\log(1+k/n)}{k}}.$$ It happens that $$\sum_{k=1}^n\frac{\log(1+k/n)}{k}\leq\int_{0}^{1}\frac{\log(1+x)}{x}dx=\frac{\zeta(2)}{2}=\frac{\pi^2}{12},$$ so, again: $$\left|1-\sum_{k=1}^n\frac{\log(1+k/n)}{k}\right|=O\left(\frac{1}{\log n}\right).$$

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It is more clear now, except of the last part with modulo, how was it obtain? – user142893 May 25 '14 at 15:26

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