Is it convergent?

Relevant wiki: Riemann Sums

$\begin{aligned} \lim_{n \to \infty} L(n) & = \lim_{n \to \infty} \frac 1{n+1} + \frac 1{n+2} + \frac 1{n+3} + \cdots + \frac 1{n+n} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac 1{n+k} = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \frac 1{1+\frac kn} & \small \color{#3D99F6} \text{Using Riemann sums: } \\ & = \int_0^1 \frac 1{1+x} dx = \ln (1+x) \bigg|_0^1 = \ln 2 \approx \boxed{0.693} & \small \color{#3D99F6} \lim_{n\to\infty} \frac 1n \sum_{k=a}^b f\left(\frac kn\right) = \lim_{n \to \infty} \int_{a/n}^{b/n} f(x)\ dx \end{aligned}$

Here is another way of doing it.

$\displaystyle \dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2n}-\left ( 1+\dfrac{1}{2}+...+\dfrac{1}{n} \right )$

Now, the second term(the one in parentheses) can be re-written as

$\displaystyle \dfrac{2}{2}+\dfrac{2}{4}+...+\dfrac{2}{2n}$

So, the whole sum becomes

$\displaystyle 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n}$

And clearly, the limit is $\displaystyle \ln{2} \approx 0.693$ as can be shown using series for $\displaystyle \ln(1+x)$ with $\displaystyle |x|<1$