I'm trying to figure out if I can interpret this riemann sum
$$ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{2n+k} $$
as
$$ \lim_{x \to \infty} \int_1^x \frac{1}{2x+y}dy $$
It seems it is correct because both expression give me the same result, $\ln\dfrac{3}{2}$, but I don't see how to derive the second expression from the first one.
Thanks in advance,
Best regards.
Tom.
Hint: Write the series as: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{1}{2+k/n}$$ now the general trick is to replace $\frac{1}{n}$ with $dx$ and determine the limits of integration, which are: $$\int_0^1\frac{1}{2+x}dx$$ Edit:
To imply your integral from that infinite sum, first note that: $$\lim_{x\to\infty}\int_1^x\frac{1}{2x+y}dy=\lim_{n\to\infty}\int_1^n\frac{1}{2n+x}dx$$ Now take a look at the picture in this post to see: $$\sum_{k=2}^n\frac{1}{2n+k}<\int_1^n\frac{1}{2n+x}dx<\sum_{k=1}^{n-1}\frac{1}{2n+k}$$ Hence regarding: $$\lim_{n\to\infty}\sum_{k=2}^n\frac{1}{2n+k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{2n+k}=\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{1}{2n+k}$$ and the sandwich theorem, proves the statement.