Evaluation of the limit $\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)$
Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$
I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.
Any help would be greatly appreciated, thanks! Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :)
Note that $$ 2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k+1}+\sqrt{k}}\leq\frac{1}{\sqrt{k}}\leq \frac{2}{\sqrt{k}+\sqrt{k-1}}=2(\sqrt{k}-\sqrt{k-1}) $$ Hence $$ 2(\sqrt{n+1}-1)=\sum\limits_{k=1}^n 2(\sqrt{k+1}-\sqrt{k})\leq\sum\limits_{k=1}^n\frac{1}{\sqrt{k}}\leq \sum\limits_{k=1}^n 2(\sqrt{k}-\sqrt{k-1})=2\sqrt{n} $$ so $$ \frac{2\sqrt{1 + n}-2}{\sqrt{n}}\leq\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\leq2 $$ The rest is clear.
Rewrite as $$\frac 1n \left(\frac 1 {\sqrt{\frac 1n}}+\frac 1 {\sqrt{\frac 2n}}+\dots +\frac 1 {\sqrt{\frac nn}} \right)$$ and interpret this as a Riemann sum for the function $$\frac 1{\sqrt x}.$$
This is a standard Stolz-Cezaro problem
$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_{n \to \infty } \frac{\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)}{\sqrt{n}}=$$ $$=\lim_{n}\frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}$$
rationalize the denominator you get
$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_n \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}=2$$
To use as little machinery as possible, observe that $\left(\sqrt n+\frac1{2\sqrt n}\right)^2=n+1+\frac1{4n}>n+1$, hence $$\tag1\sqrt{n+1}<\sqrt n + \frac1{2\sqrt n}.$$ By induction, we see therefore that $$\tag22\sqrt {n+1}<2+\sum_{k=1}^n \frac1{\sqrt k}\quad\text{for all }n\in\mathbb N.$$ On the other hand, if $q>2$, then for $n$ sufficiently large, we have $\left(\sqrt n+\frac1{q\sqrt n}\right)^2=n+\frac2q+\frac1{q^2n}<n+1$ and hence $$\tag3\sqrt{n+1}>\sqrt n + \frac1{q\sqrt n}.$$ Again by induction, we therefore find $$\tag4q\sqrt {n+1}>C+\sum_{k=1}^n \frac1{\sqrt k}\quad\text{for all }n\in\mathbb N,$$ where $C$ is a (negative) constant depending on $q$ (needed to cover the fact that $(3)$ hold only for $n$ sufficiently large). This gives us $$\frac{2\sqrt{n+1}-2}{\sqrt n}<\frac1{\sqrt n}\left(1+\frac1{\sqrt 2}+\cdots+\frac1{\sqrt n}\right)<\frac{q\sqrt{n+1}-C}{\sqrt n}$$ for almost all $n$. The left and right estimate converge to $2$ and $q$, respectively, as $n\to\infty$. Since $q$ was any number $>2$, we conclude that $$\lim_{n\to\infty}\frac1{\sqrt n}\left(1+\frac1{\sqrt 2}+\cdots+\frac1{\sqrt n}\right)=2.$$
A couple of other methods, which use more machinery than the earlier ones, but are worth looking into for their generality.
Bounding by integrals
Comparing the integrals by constant bounding function over unit intervals, we get $$ \frac1{\sqrt{n}}\left(\int_1^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\right) \le\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}} \le\frac1{\sqrt{n}}\left(1+\int_1^n\frac{\mathrm{d}x}{\sqrt{x}}\right) $$ which yields $$ \color{#C00000}{\frac1{\sqrt{n}}\left(2\sqrt{n+1}-2\right)} \le\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}} \le\color{#C00000}{\frac1{\sqrt{n}}\left(2\sqrt{n}-1\right)} $$ As $n\to\infty$, both bounding terms (in red) tend to $2$. Therefore, by the Squeeze Theorem, we get $$ \lim_{n\to\infty}\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}}=2 $$ Euler-Maclaurin Sum Formula
The Euler-Maclaurin Sum Formula says that for some constant $C$, we have $$ \sum_{k=1}^n\frac1{\sqrt k}=2\sqrt{n}+\frac1{2\sqrt n}+C+O\left(n^{-3/2}\right) $$ Dividing by $\sqrt n$ yields $$ \begin{align} \frac1{\sqrt n}\sum_{k=1}^n\frac1{\sqrt k} &=2+\frac C{\sqrt n}+\frac1{2n}+O\left(n^{-2}\right)\\ &\to2 \end{align} $$