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I have been trying to find the limit of the following sequence:

$\lim_{n\to \infty}(\frac{1}{2n+1} + \frac{1}{2n+2}+...+\frac{1}{8n+1})$

Here is my attempt with the Euler-Mascheroni constant:

$(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n+1} + \frac{1}{2n+2}+...+\frac{1}{8n+1}) - (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}) =$

$=\gamma + \varepsilon_{8n+1} + \ln(8n+1) -(\gamma + \epsilon_{2n} + \ln 2n)$

$=\ln\frac{8n+1}{2n}$

Can someone check if this is correct or is there another way to solve this problem?

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Yes this method is a correct approach. As we take the limit the result is then given by $$\lim_{n \to \infty} \ln{\frac{8n+1}{2n}}=2\ln{2}$$

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I think you can also treat this as a Riemann sum:

$$\sum_{k=1}^{6n+1} \frac{1}{2n+k}=\frac{1}{n}\sum_{k=1}^{6n+1} \frac{1}{2+\frac{k}{n}}\longrightarrow \int_0^{6}\frac{1}{2+x}\,dx=\ln 4$$

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