<s>[INST] Sequence converging towards Euler-Mascheroni constant 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?
 [/INST] 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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