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

My work
$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\frac{\sum_{m=1}^n\frac{\sqrt m}{m+1}}{\sqrt n}$$
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?
 A: $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt r}{(r+1)\sqrt n}$$
$$=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{n\frac{(r+1)}{n}}$$
$$=\lim_{n\rightarrow\infty }\frac 1n\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{\frac rn + \frac1n}$$
$$=\int\limits_0^1\frac{\sqrt x}{dx+x}dx \approx \int\limits_0^1\frac{\sqrt x}{x}dx$$
$$=\boxed{2}$$
A: 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
$$
\frac{a_n}{b_n}=\frac{\frac{\sqrt{n}}{n+1}}{\sqrt{n}-\sqrt{n-1}}=\frac{n+\sqrt{n(n-1)}}{n+1}\xrightarrow{n\to \infty} 2.
$$
A: 
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:
$$\boxed{\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n} = 2}$$
A: The fact that $$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$ diverges does not mean $$\sum_{m=1}^n \frac{\sqrt m}{m+1}$$ does too.
For instance, the harmonic sum $\sum_{m=1}^\infty \frac{1}{m}$ is divergent, but the sum up to $n$, i.e., $\sum_{m=1}^n \frac{1}{m}$ is a rational number.
