Evaluating $\lim_{n\to \infty}\frac{1+\sqrt{2}+\dots+\sqrt{n}}{n\sqrt{n}}$ using inequalities

Question

Compute: $$L=\lim_{n\to \infty}\frac{1+\sqrt{2}+\dots+\sqrt{n}}{n\sqrt{n}}$$

I know some calculation of this limit using integrals, the anwser is $L=\frac{2}{3}$. However, I have an idea to compute it simpler by using just simple inequalities. My idea is to find an upper-bound for $1+\sqrt{2}+\dots+\sqrt{n}$ of the form $\frac{2}{3}n\sqrt{n+c}$ for some constant $c>0$ and its lower-bound is just $\frac{2}{3}n\sqrt{n}$. Hence, we got $2/3\leq L\leq \lim_{n\to \infty}\frac{2/3n\sqrt{n+c}}{n\sqrt{n}}=2/3$.

1. For lower-bound: I can simply prove $1+\sqrt{2}+\dots+\sqrt{n}\geq \frac{2}{3}n\sqrt{n}$ for every $n$ by induction.
2. For upper-bound: I'm stucking at choosing $c$, I think $c=7$ works.

How can I continue this direction?

Answer 1

My claim : $\frac{2}{3}(n+1)\sqrt{n+1} \gt \sum_{r=1}^n \sqrt{r}$
To prove this we shall use induction, for $n=1$ this is true.
Now assume this be true for $n-1$. So we have :
$\frac{2}{3}n\sqrt{n} \gt \sum_{r=1}^{n-1}\sqrt{r}$
$ \Rightarrow \frac{2}{3} n\sqrt{n} +\sqrt{n}\gt \sum_{r=1}^n \sqrt{r}$
So if we show $\frac{2}{3}(n+1)\sqrt{n+1} \gt \frac{2}{3} n\sqrt{n} +\sqrt{n}$ we are done
$\frac{2}{3}(n+1)\sqrt{n+1} - \frac{2}{3} n\sqrt{n} -\sqrt{n}$
$ = \frac{2}{3} \left( (n+1)^{\frac{3}{2}} - (n)^{\frac{3}{2}}\right)-\sqrt{n}$
$ = \frac{2}{3}\left( \sqrt{n+1} - \sqrt{n} \right) \left( n+1 + n + \sqrt{n(n+1)} \right) -\sqrt{n}( \sqrt{n+1} - \sqrt{n} )( \sqrt{n+1} + \sqrt{n} )$
$ = \frac{2}{3}\left( \sqrt{n+1} - \sqrt{n} \right) \left( n+1 + n + \sqrt{n(n+1)} -\frac{3}{2}n -\frac{3}{2}\sqrt{n(n+1)}\right) $
$ = \frac{2}{3}\left( \sqrt{n+1} - \sqrt{n} \right) \left( \frac{1}{2} +\frac{n+1}{2} -\frac{1}{2} \sqrt{n(n+1)} \right)$
$ =\frac{1}{3}\left( \sqrt{n+1} - \sqrt{n} \right) \left(1+ \sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n} \right)\right) \gt 0$
So we have proved for $n$.
Thus we have : $$\frac{2}{3}(n+1)\sqrt{n+1} \gt \sum_{r=1}^n \sqrt{r} \gt \frac{2}{3}n\sqrt{n}$$

Answer 2

Remark. We can use the method of unknown coefficients (by assuming the form of the upper and lower bounds) and the Mathematical Induction to evaluate the limit.

First of all, we need to guess the appropriate form of the upper and lower bounds. Here we guess that there exist constants $L$ and $c_1, c_2$ such that for sufficiently large $n$, $$L\cdot n\sqrt{n} - c_1 n\le 1 + \sqrt{2} + \cdots + \sqrt{n} \le L \cdot n\sqrt{n} + c_2 n.$$ (Note: We can guess other forms. For example, if we replace $c_1n$ and $c_2n$ with $c_1\sqrt{n}$ and $c_2\sqrt{n}$ respectively, it still works. )

If the Mathematical Induction works, we have \begin{align*} 1 + \sqrt{2} + \cdots + \sqrt{n} + \sqrt{n+1} &\le L\cdot n \sqrt{n} + c_2 n + \sqrt{n+1} \\ &\le L \cdot (n+1)\sqrt{n+1} + c_2(n+1), \tag{1} \end{align*} and \begin{align*} 1 + \sqrt{2} + \cdots + \sqrt{n} + \sqrt{n+1} &\ge L\cdot n\sqrt{n} - c_1 n + \sqrt{n+1}\\ &\ge L\cdot (n+1)\sqrt{n+1} - c_1 (n+1). \tag{2} \end{align*}

From (1), we have $$L[(n+1)^{3/2} - n^{3/2}] + c_2 - \sqrt{n+1} \ge 0,$$ or $$L[(n+1) - n^{3/2}(n+1)^{-1/2}] + \frac{c_2}{\sqrt{n+1}} - 1 \ge 0. \tag{3}$$ Taking limit on (3), using $\lim_{n\to \infty} [(n+1) - n^{3/2}(n+1)^{-1/2}] = \frac32$, we have $\frac32 L - 1 \ge 0$. Thus, $L \ge \frac23$.

From (2), we have $$c_1 + \sqrt{n+1} - L[(n+1)^{3/2} - n^{3/2}] \ge 0,$$ or $$\frac{c_1}{\sqrt{n+1}} + 1 - L[(n+1) - n^{3/2}(n+1)^{-1/2}] \ge 0.\tag{4}$$ Taking limit on (4), using $\lim_{n\to \infty} [(n+1) - n^{3/2}(n+1)^{-1/2}] = \frac32$, we have $1 - \frac32 L \ge 0$. Thus, $L \le \frac23$.

Thus, we have $L = \frac23$.

The rest is smooth. For example, we choose $c_1= c_2 = 1$, and indeed the Mathematical Induction works.