TITLE: Finding $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: 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
$$
\frac{a_n}{b_n}=\frac{\frac{\sqrt{n}}{n+1}}{\sqrt{n}-\sqrt{n-1}}=\frac{n+\sqrt{n(n-1)}}{n+1}\xrightarrow{n\to \infty} 2.
$$