Evaluating two limits as limits of Riemann sums. I mention Riemann sums because they appear on the chapter on Riemann integrals. But I don't see how the can be expressed in a computable form, they are:
$1) \displaystyle \lim _{n\to \infty }\sqrt[n]{\left(1+\frac{2}{n}\right)...\left(1+\frac{2n}{n}\right)}$
$2) \displaystyle \lim _{n\to \infty }\sum _{k=qn}^{pn}\frac{1}{k }, \ p,q \in \mathbb{N}^* , \ p > q \ge 1$
The solutions are as given:
$1) \displaystyle \ \frac{3\sqrt{3}}{e} $
$2) \displaystyle \log \frac{p}{q}$

I've got nothing for $1)$, I've managed to bound it between $1$ and $3$ but I can't figure how to turn it into a sum.
From the result of $2)$ it's pretty clear that the sum converges to $\displaystyle \int^p_q \frac{\mathrm{d}x}{x}$. So we would have $f(x) = \frac{1}{x}$ on the interval $[q,p]$, but I don't see what partition to use, and I guess the point's on each partition must be $\xi_k=x_k $, with $[x_{k-1},x_k]$ being the partitions of $[q,p]$. I was thinking of splitting the sum into a difference of sums, each from $k=1$ to $pn$ and $qn$ respectively. 
Thanks!
 A: For 1), apply the logarithm first, to turn the product into a sum.
For 2), do the obvious thing, i.e. partition on the natural numbers, $[n,n+1)$ and give it some more thought.
A: The first problem is a little ambiguous. I will assume that the product is supposed to be $(1+2/n)(1+4/n)(1+6/n)\cdots(1+2n/n)$.
Take the logarithm. We get 
$$\frac{1}{2}\left[\frac{2}{n}\left(\ln(1+2/n)+\ln(1+4/n)+\cdots +\ln(1+2n/n)  \right)\right].\tag{1}$$ 
The sum
$$\frac{2}{n}\left(\ln(1+2/n)+\ln(1+4/n)+\cdots +\ln(1+2n/n)  \right)\tag{2}$$
can be recognized as a right Riemann sum for the integral $\int_0^2 \ln(1+x)\,dx$ (the partition size is $\frac{2}{n}$). 
Integrate. An antiderivative is $(1+x)\ln(1+x)-x$, so the definite integral is $3\ln 3-2$.  That is the limit of (2), and therefore (1) has limit $\frac{3\ln 3}{2}-1$. It follows that our original limit is
$$\exp\left(\frac{3\ln 3}{2}-1\right),$$
which simplifies to $3^{3/2}/e$.
A: 2) Let $S_n$ be the $n$th sum. Look at rectangles of base $1$ and heights along the graph of $1/x.$ Because $1/x$ decreases, we get the inequality
$$\int_{qn}^{pn+1}\frac{dx}{x} \le S_n \le \frac{1}{qn} + \int_{qn}^{pn}\frac{dx}{x}.$$
Integrating then gives $\ln[(pn+1)/qn] \le S_n \le 1/(qn) + \ln[(pn)/qn].$ Now let $n\to \infty$ and apply the squeeze theorem.
