find the limit of the sum $\sum_{k=n}^{2n-1}\frac{1}{2k+1}$ Find
$$\lim_{n\to+\infty}\sum_{k=n}^{2n-1}\frac{1}{2k+1}$$
i tried to write as
$$\sum_{k=0}^{n-1}\frac{1}{2k+2n+1}$$
$$=\frac 1n\sum_{k=0}^{n-1}\frac{1}{\frac{2k+1}{n}+2}$$
like a Riemann sum, but i can't find the corresponding function.
any idea will be appreciated.
 A: \begin{align*}
\dfrac{1}{n}\sum_{k=0}^{n-1}\dfrac{1}{\dfrac{2(k+1)}{n}+2}\leq\dfrac{1}{n}\sum_{k=0}^{n-1}\dfrac{1}{\dfrac{2k+1}{n}+2}\leq\dfrac{1}{n}\sum_{k=0}^{n-1}\dfrac{1}{\dfrac{2k}{n}+2}.
\end{align*}
The left and right sided tend to
\begin{align*}
\int_{0}^{1}\dfrac{1}{2x+2}dx.
\end{align*}
A: As a general approach, sums like this often can be evaluated as a double limit
$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{n}} = \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{m}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 }= \int_0^2 \frac{dx}{2x+2}$$
The first step, $\lim_{n \to \infty} S_{nn} = \lim_{n\to \infty} \lim_{m \to \infty} S_{mn}$ is valid for a double sequence $(S_{mn})$ when we have  $S_{mn} \to T_n$ as $m \to \infty$ uniformly for all $n$.
In this case, convergence  is uniform as $m \to \infty$ for all $n$, since
$$\left| \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 + \frac{1}{m}}-  \frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{2k}{n} + 2 } \right| = \left| \frac{1}{n}\sum_{k=0}^{n-1}\frac{-\frac{1}{m}}{\left(\frac{2k}{n} + 2 + \frac{1}{m}\right)\left(\frac{2k}{n} + 2 \right)}\right| \\ \leqslant  \frac{1}{nm}\sum_{k=0}^{n-1}\frac{1}{\left|\frac{2k}{n} + 2 + \frac{1}{m}\right|\left|\frac{2k}{n} + 2 \right|}\leqslant \frac{1}{nm}\sum_{k=0}^{n-1}\frac{1}{4} = \frac{1}{4m}\underset{m\to \infty}\longrightarrow 0$$
