Calculate limit using Riemann integral

Some of the limits are easy to see, but I want to use Riemman integral to determine the value of the following limits. $$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n})$$ $$\lim_{n\to\infty} \frac{1}{n} \sqrt[n]{(n+1)(n+2)\cdots(n+n)}$$ $$\lim_{n\to\infty} (\sin {\frac{n}{n^2+1^2}}+\sin {\frac{n}{n^2+2^2}}+\cdots+\sin {\frac{n}{n^2+n^2}})$$

For exmaple in first, I suppose $\int_a^b f(x)dx$ is equal to the limit $$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}) = \lim_{n\to\infty} \sum_{i = 1}^{2n} \frac{1}{n+i} = \lim_{n\to\infty}\sum_{i = 1}^{2n} f(a+\frac{(b-a)i}{2n})\cdot\frac{(b-a)i}{n}$$ But cannot find anything.

Please don't tell me the method directly, I would appreciate it if you could give me some crucial hints for one\all of them. Thank you very much.

Solution 1:

For the second $ \displaystyle \lim_{n\to\infty} \frac{1}{n} \sqrt[n]{(n+1)(n+2)\cdots(n+n)} = \lim_{n\to \infty} \left( \left( 1 + \frac 1 n \right)\left( 1 + \frac 2 n \right) ... \left( 1 + \frac{n}{n}\right)\right)^{\frac 1 n}$ which is equal to $\displaystyle \lim_{n\to \infty } e^{ \frac 1 n \sum_{k=1}^n\log \left( 1 + \frac k n \right)}$

Solution 2:

For the first, rewrite $\dfrac{1}{n+k}$ as $\dfrac{1}{n}\dfrac{1}{1+\frac{k}{n}}$, and bring the common $\dfrac{1}{n}$ in front.

Kind of looks like a Riemann sum, step size $\dfrac{1}{n}$, integrating $\dots$ from $0$ to $\dots$. (I could fill in the $\dots$, but so can you.)