How to prove that $\lim_{n \to \infty} \frac{1}{\sqrt{n}} (-\frac{2}{3} n^{\frac{3}{2}} + \sum^{n}_{k=1} \sqrt{k})$ is equal to one half?

Question

I was trying to construct a polynomial that was approximately equal to $\sum^{x}_{k=1} \sqrt{k}$. I knew the first term must be $\frac{2}{3} x^{3/2}$ and the second term must be $\sqrt{x}$ times some constant, $a$. Below I solved, for $a$.

$$\sum^{x}_{k=1} \sqrt{k} \approx \frac{2}{3} x^{3/2}+ a \sqrt{x}$$

$$a \approx \frac{-\frac{2}{3} x^{3/2} + \sum^{x}_{k=1} \sqrt{k}}{\sqrt{x}}$$

$$a = \lim_{n \to \infty} \frac{-\frac{2}{3} n^{3/2} + \sum^{n}_{k=1} \sqrt{k}}{\sqrt{n}}$$

$$a=\frac{1}{2}$$

My issue is that I know no other means to calculate the value of $a$ without plugging in huge numbers and guessing. Does anyone know how to prove the limit more rigorously?

Answer 1

Rewriting it as $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \left(-\frac{2}{3} n + \sum^{n}_{k=1} \sqrt{k}\right) = \lim_{n\to\infty} n\left(\color{red}{\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}}} - \frac{2}{3\sqrt n}\right)$$ and noting that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}} = \int_0^1 \sqrt x\mathrm dx = \color{red}{\frac{2}{3}}$$ The limit seems to be $\infty$.

Answer 2

I am getting infinity, here is my method,

Let, $$P = \lim_{n\to \infty}{\frac{1}{\sqrt{n}}\biggl(\frac{-2}{3}n+\sum_{k=1}^{n}\sqrt{k}\biggr)}$$ $$= \lim_{n\to \infty}\frac{-2}{3}\sqrt{n}+\lim_{n\to \infty}n\biggl(\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n}\sqrt{\frac{k}{n}}\biggr)$$ $$= \lim_{n\to \infty}\frac{-2}{3}\sqrt{n}+\lim_{n\to \infty}n\biggl(\int_{0}^{1}\sqrt{x}dx\biggr)$$ $$= \lim_{n\to \infty}\frac{-2}{3}\sqrt{n}+\lim_{n\to \infty}\frac{2}{3}n$$ So, $P {\to \infty}$, as $n{\to \infty}$.

Answer 3

$$\sqrt{x-1}\le\sqrt{\lfloor x\rfloor}\le\sqrt x$$

and by integration from $1$ to $n+1$,

$$\int_1^{n+1}\sqrt{x-1}\,dx<\sum_{k=1}^n\sqrt k<\int_1^{n+1}\sqrt x\,dx$$ or

$$\frac23n^{3/2}<\sum_{k=1}^n\sqrt k<\frac23((n+1)^{3/2}-1)\sim\frac23\left(n^{3/2}\left(1+\frac3{2n}+o\left(\frac1n\right)\right)-1\right).$$

From this, we can conclude that

$$0<\frac{\displaystyle\sum_{k=1}^n\sqrt k-\dfrac23n^{3/2}}{\sqrt n}<1.$$

The exact value can be obtained by the Euler summation formula.

Answer 4

$$\frac1{\sqrt n}\left(-\frac23n+\sum_{k=1}^n\sqrt n\right)\ge\frac1{\sqrt n}\left(-\frac23n+ n\right)=\frac13\sqrt n\xrightarrow[n\to\infty]{}\infty$$

It must be that Wolfram is high again...as in many other cases.

Answer 5

Note that we can proceed using the Euler-Maclaurin Summation Formula, but this is unnecessary machinery here.

Alternatively, using creative telescoping, we see that

$$\frac23\sum_{k=1}^n \left(k^{3/2}-(k-1)^{3/2}+\frac34 \left(k^{1/2}-(k-1)^{1/2}\right)\right)=\frac23n^{3/2}+\frac12 n^{1/2}\tag1$$

and

$$\frac23\left(k^{3/2}-(k-1)^{3/2}+\frac34 \left(k^{1/2}-(k-1)^{1/2}\right)\right)=\sqrt{k}+O\left(\frac1{k^{3/2}}\right)\tag2$$

Hence, using $(1)$ and $(2)$, we can assert that

$$\sum_{k=1}^n \sqrt{k}=\frac23n^{3/2}+\frac12 n^{1/2}+\underbrace{\sum_{k=1}^n O\left(\frac1{k^{3/2}}\right)}_{\text{converges}}\tag3$$

Finally, rearranging $(3)$ reveals

$$\frac1{\sqrt n}\left(-\frac23 n^{3/2}+\sum_{k=1}^n \sqrt{k}\right)=\frac12+\underbrace{\frac1{\sqrt n}\sum_{k=1}^n O\left(\frac1{k^{3/2}}\right)}_{\to 0\,\,\text{as}\,\,n\to \infty}$$

from which we obtain the coveted limit

$$\lim_{n\to\infty}\frac1{\sqrt n}\left(-\frac23 n^{3/2}+\sum_{k=1}^n \sqrt{k}\right)=\frac12$$

as was to be shown!