$lim_{n \to \infty} [\frac1{n+1} + \frac1{n+2} + ..... + \frac1{kn}]=?$, where $k \gt 1$ is an integer.

Question

Suppose $k$ is an integer greater than 1.

What is the value of $$lim_{n \to \infty} [\frac1{n+1} + \frac1{n+2} + ..... + \frac1{kn}]?$$

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Is it $0$ as is apparent, or is there some way to manipulate the expression otherwise?

Answer 1

$$\lim_{n\to\infty}\sum_1^{(k-1)n}\frac{1}{n+r}=\lim_{n\to\infty}\sum_1^{(k-1)n}\frac{1/n}{1+r/n}=\int_0^{k-1}\frac{dx}{1+x}$$

Note that infinite sum of $0^+$ may not tend to zero as in this case.

Answer 2

Since we have the next limit: $$a_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\log{n}\to\gamma$$ as $n\to\infty$, where $\gamma$ is the Euler constant.

Then we have $$\frac{1}{n+1}+\ldots+\frac{1}{kn}=a_{kn}-a_{k}+\log{(kn)}-\log{n} =a_{kn}-a_{k}+\log{k}\to 0$$ as $n\to\infty$.

Thus we obtain that $$\lim_{n\to \infty}\frac{1}{n+1}+\ldots+\frac{1}{kn}=\log{k}.$$

Answer 3

Using harmonic numbers $$S_n=\sum_{r=1}^{(k-1)n}\frac{1}{n+r}=H_{k n}-H_n$$ Using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ we then obtain $$S_n=\log (k)-\frac{1-\frac{1}{k}}{2 n}+\frac{1-\frac{1}{k^2}}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.