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First I had to calculate

$$\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n(n+k)}$$

So, if $k>m$, then $$\sum_{n=1}^\infty\frac{k}{n(k+n)}>\sum_{n=1}^m\frac{m}{2mn}=\frac12H_m\;,$$

where $H_m=\sum_{k=1}^m\frac1k$ is the $m$-th harmonic number. (And we know that the harmonic series diverges)

But now I have to show with the above that

$$\sum_{k=1}^n \frac{1}{n+k} \to \log 2 \text{ for } n \to \infty$$

Can someone tell me how one can prove that with the info given above

stepbysteptomathpro
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    See this old thread for the last question. My answer is less relevant to you, but the others describe the connection to harmonic numbers. Overall, the quality of answers in that thread is high in comparison. – Jyrki Lahtonen Nov 28 '19 at 05:43

1 Answers1

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We have that

$$\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n(n+k)}=\lim_{n\to \infty} \frac1n\sum_{k=1}^n \frac{\frac kn}{1+\frac kn}=\int_0^1\frac{x}{1+x}dx=1-\log 2$$

and

$$\sum_{k=1}^n \frac{k}{n(n+k)}=\sum_{k=1}^n \frac{1}{n}-\sum_{k=1}^n \frac{1}{n+k}= 1-\sum_{k=1}^n \frac{1}{n+k}$$

user
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