How can i evaluate $\lim_{n\to\infty}\left[ \left(\frac{1}{n}\right)^\frac{3}{2} \sum\limits_{i=2}^{n+1} i^{\frac{1}{2}}\right] $

Question

I've not looked at finding the limits of sums before, and i'm interested to know how this particular example would be evaluated. The answer I believe should give $\frac{2}{3}$.

Answer 1

The usual way: $\sqrt{x}$ is a continuous function on $[0,1]$, hence it is Riemann-integrable and $$\int_{0}^{1}\sqrt{x}\,dx = \lim_{n\to+\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}}\tag{1}$$ The RHS of $(1)$ is your limit and by the fundamental theorem of Calculus the LHS of $(1)$ is $\frac{1}{1+\frac{1}{2}}=\color{red}{\large\frac{2}{3}}$ as wanted.

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The unusual way: it is enough to understand the behaviour of $\sum_{k=1}^{n}\sqrt{k}$ with decent accuracy.
For such a task, we may employ creative telescoping, since

$$ \left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}-\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}} = \frac{3k^2+\frac{1}{4}}{\left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}+\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}}}$$ differs from $\frac{3}{2}\sqrt{k}$ by less than $\frac{1}{60 k^{3/2}}$ for any $k\geq 1$. In particular, $$ \left|\sum_{k=1}^{n}\sqrt{k}-\frac{2}{3}\left(n+\frac{1}{2}\right)^{3/2}\right|\leq \frac{1}{3\sqrt{2}}+\frac{1}{90}\sum_{k\geq 1}\frac{1}{k^{3/2}} = C $$ and the wanted limit is $\color{red}{\large\frac{2}{3}}$.

Answer 2

METHODOLOGY $1$:

Note that since $\sqrt x$ is monotonically increasing we can bound the sum of interest $S_n=\sum_{i=2}^{n+1}i^{1/2}$ by

$$\int_1^{n+1}\sqrt{x}\,dx\le S_n=\sum_{i=2}^{n+1}i^{1/2}\le \int_2^{n+2}\sqrt x\,dx$$

whereupon evaluating the integrals gives

$$\frac23((n+1)^{3/2}-1)\le \sum_{i=2}^{n+1}i^{1/2}\le \frac23((n+2)^{3/2}-2^{3/2})\tag 1$$

Dividing $(1)$ by $n^{3/2}$ and applying the squeeze theorem yields the coveted result

$$\lim_{n\to \infty}\frac{S_n}{n^{3/2}}=\frac23$$

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METHODOLOGY $2$:

This is overkill for the purpose herein, but I thought it might be instructive to present an approach that provides an expansion of $S_n=\sum_{i=2}^{n+1}i^{1/2}$. To that end we proceed.

Using the Euler-Maclaurin Summation Formula, we can write

$$\begin{align} S_n&=\sum_{i=2}^{n+1}i^{1/2}\\\\ &=\int_1^{n+1}\sqrt{x}\,dx+\frac12(\sqrt{n+1}-\sqrt{1})+\frac{B_2}{2!}\frac12\left((n+1)^{-1/2}-^{-1/2}\right)\\\\ &+\frac{B_4}{4!}\frac{3}{8}\left((n+1)^{-5/2}-(1)^{-5/2}\right)+C+O\left((n+1)^{-9/2}\right)\\\\ &=\frac23(n+1)^{3/2}+\frac{\sqrt{n+1}}{2}+C'+\frac{1}{24}(n+1)^{-1/2}-\frac{1}{1920}(n+1)^{-5/2}+O\left((n+1)^{-9/2}\right) \end{align}$$

The constant $C'$ is given by the limit

$$\begin{align} C'&=\lim_{n\to \infty}\left(\sum_{i=2}^{n+1}i^{1/2}-\frac23(n+1)^{3/2}-\frac{\sqrt{n+1}}{2}\right)\\\\ &=\zeta\left(-\frac12\right)\\\\ &\approx 0.207886225 \end{align}$$

Hence, we have

$$\begin{align} S_n=\frac23(n+1)^{3/2}+\frac{\sqrt{n+1}}{2}+\zeta\left(-\frac12\right)+\frac{(n+1)^{-1/2}}{24}-\frac{(n+1)^{-5/2}}{1920}+O\left((n+1)^{-9/2}\right) \tag 2 \end{align}$$

Now, we see from $(2)$ that $\lim_{n\to \infty}\frac{S_n}{n^{3/2}}=\frac23$.

Answer 3

Hint: $$\ \frac1{n^{\frac32}}\sum_{i=\color{red}1}^{\color{red}n}=\frac1n \sum_{i=1}^{n}\Bigl(\frac in\Bigr)^{\!\frac12}$$ is an upper Riemann sum for the integral of a function on the interval $[0,1]$.