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i have a question about this limit.

$\lim\limits_{n\to \infty } \ (\frac{1}{3n} + \frac{1}{3n+1} + \frac{1}{3n+2} + \dots + \frac{1}{4n} )$

I tried to calculate it using the squeeze theorem and got $L = \frac{1}{4}$ (i'm not sure that its correct).

Also i have a feeling that maybe it should be done with riemann sums , but i couldnt manage to solve it like that.

if anyone can show me how to find the limit it would be great!

thanks in advance!

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We have

$$\frac{1}{3n} + \frac{1}{3n+1} + \frac{1}{3n + 2} + \cdots + \frac{1}{4n} = \frac{1}{3n} + \frac{1}{n}\left(\frac{1}{3 + \frac{1}{n}} + \frac{1}{3 + \frac{2}{n}} + \cdots + \frac{1}{3 + \frac{n}{n}}\right),$$

and $\frac{1}{3n} \to 0$. So the limit of your sequence is the same as the limit of the sequence on the right, which is a sequence of Riemann sums of $f(x) = \frac{1}{3 + x}$ over the interval $[0,1]$. So it converges to

$$\int_0^1 \frac{1}{3 + x}\,dx = \log|3 + x|\bigg|_{x = 0}^1 = \log(4) - \log(3) = \log(4/3).$$

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$$\lim_{n\to\infty}\sum_{r=1}^n\frac1{3n+r}=\lim_{n\to\infty}\frac1n\sum_{r=1}^n\frac1{3+r/n}$$

Now see The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

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The sum is equal to a difference between $H_{4n}$ and $H_{3n}$, where $H_n \sim \log n + \gamma + o(1)$.

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