Finding $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$

The series $$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$ does not converge so can I say $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$ does not exist?

Because Zacky deleted his answer, I'll repeat the useful observation that divergence of a numerator does not mean the fraction necessarily diverges...

As for finding the limit; you can rewrite towards a Riemann sum: $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{k=1}^{n}\frac{\sqrt k}{(k+1)\sqrt n}=\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}} \tag{$\star$}$$ Now you have an upper bound: $$(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}}\color{blue}{\le}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k}{n}}}} = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{blue}{2}$$ but also a lower bound: $$\begin{align}(\star) : \lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\frac{k+1}{\sqrt {kn}}} =\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt\frac{k^2+2k+1}{{kn}}} & \color{red}{\ge}\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{k=1}^{n}\frac{1}{{\sqrt {\frac{k+3}{n}}}}\\ & =\lim_{n\rightarrow\infty }\frac{1}{n}\sum_{m=4}^{n+3}\frac{1}{{\sqrt {\frac{m}{n}}}}\\[5pt] & = \int_0^1 \frac{1}{\sqrt{x}}\,\mbox{d}x = \color{red}{2}\end{align}$$ So we have: $$\boxed{\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n} = 2}$$

$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt r}{(r+1)\sqrt n}$$

$$=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{n\frac{(r+1)}{n}}$$ $$=\lim_{n\rightarrow\infty }\frac 1n\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{\frac rn + \frac1n}$$

$$=\int\limits_0^1\frac{\sqrt x}{dx+x}dx \approx \int\limits_0^1\frac{\sqrt x}{x}dx$$ $$=\boxed{2}$$

By the theorem of Cesàro-Stolz (a discrete version of the l'Hopital rule for $\frac\infty\infty$), the quotient $\frac{\sum_{m=1}^n a_m}{\sum_{m=1}^n b_m}$ has a limit if the denominator grows to infinity and the quotient $\frac{a_n}{b_n}$ of the last terms has a limit, and then both limits have the same value.

Here $$ \frac{a_n}{b_n}=\frac{\frac{\sqrt{n}}{n+1}}{\sqrt{n}-\sqrt{n-1}}=\frac{n+\sqrt{n(n-1)}}{n+1}\xrightarrow{n\to \infty} 2. $$