A calculus problem by Sutirtha Datta

Calculus Level 3

$\large \lim_{n\to\infty} \dfrac{ \sqrt1 + \sqrt2 + \cdots + \sqrt{n-1}}{n \sqrt n} = \frac ab$

The equation above holds true for coprime positive integers $a$ and $b$ . Find $a+b$ .

The answer is 5.

4 solutions

Chew-Seong Cheong
Feb 4, 2017

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\sqrt 1 + \sqrt 2 + \sqrt 3 + \cdots + \sqrt {n-1}}{n \sqrt n} \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^{n-1} \sqrt {\frac kn} & \small \color{#3D99F6} \text{By Riemann's sum: } \lim_{n \to \infty} \frac 1n \sum_{k=1}^n f \left( \frac kn \right) = \int_a^b f(x) \ dx \\ & = \int_0^1 \sqrt x \ dx & \small \color{#3D99F6} a = \lim_{n \to \infty} \frac 1n = 0, \ b = \lim_{n \to \infty} \frac {n-1}n = 1 \\ & = \frac 23 x^\frac 32 \bigg|_0^1 = \frac 23 \end{aligned}$

$\implies a + b = 3 + 2 = \boxed{5}$

$L=\dfrac{2}{3}$ not $\dfrac{3}{2}$

Anirudh Sreekumar - 4 years, 4 months ago

Yes, thanks. I have amended it.

Chew-Seong Cheong - 4 years, 4 months ago

Nice observation! The key step is to recognize that $\displaystyle\sum_{k=1}^{n-1} \sqrt {\frac kn}$ is in fact equivalent to the area under the step function $f(x) = \sqrt{\frac{k}{n}}, \quad \frac{k}{n} \leq x \leq \frac{k+1}{n}$ for $k = 0, 1, 2, 3, \ldots , n-1$ .

As $n$ becomes larger, $f(x)$ becomes better and better approximation of $\sqrt{x}$ . When $n \to \infty$ , the sum approaches the area under the curve of $\sqrt{x}$ which is $\int _0 ^1 \sqrt{x} \, dx$

Pranshu Gaba - 4 years, 4 months ago

Yup, that's a very good explanation for the Riemann Sum (RS)!

For the sake of variety, one could actually solve this problem without using RS as well! We could do this by starting with Stolz–Cesàro theorem , followed by the binomial series $(1+x)^n = 1 + nx + \dfrac12 n(n-1) x^2 + \cdots$ for $|x| < 1$ .

Does anyone want to give this method a try?

Pi Han Goh - 4 years, 4 months ago

This is another great method! This is a $\infty / \infty$ case, and since the denominator is strictly increasing, we can apply the Stolz–Cesàro theorem.

$\begin{aligned} \lim_{n\to\infty} \dfrac{ \sqrt1 + \sqrt2 + \cdots + \sqrt{n-1}}{n \sqrt n} &= \lim_{n\to \infty} \frac{ \sqrt{n}}{(n+1) \sqrt {n+1} - n\sqrt{n}} \\ & = \lim_{n\to \infty} \frac{ 1}{(n+1) \sqrt {1+\frac 1n} - n}\end{aligned}$

Using the binomial series for the square root in the denominator, it tends to $\frac{3}{2}$ , and the limit of the fraction is $\frac 23$

Pranshu Gaba - 4 years, 3 months ago

Sutirtha Datta
Feb 3, 2017

$\lim_{n\rightarrow \infty}\frac{\sqrt{1}+\sqrt{2}+...+\sqrt{n-1}}{{n\sqrt{n}}}$

= $\lim_{n\rightarrow \infty}\frac{1}{n} \sum_{r=1}^{n-1}{\sqrt{\frac{r}{n}}}$

= $\int_0^1\sqrt{x}dx$

where $x$ = $\frac{r}{n}$ & $dx$ = $\frac{1}{n}$

= $\frac{2}{3}$

It would be helpful to explain what is going on.

For example, you can explain how step 2 and step 3 are equivalent because the limiting sum in Step 2 represents a Riemann sum, which is the most important part of solving this problem.

Pi Han Goh - 4 years, 4 months ago

Oleg Turcan
Feb 16, 2017

Right. To clarify, the limit can be expressed as a Riemann sum , which is the main crux of this problem.

Pi Han Goh - 4 years, 3 months ago

Noel Lo
Jul 13, 2017

$L=lim_{n\to\infty}\frac{ \sqrt1 + \sqrt2 + \cdots + \sqrt{n-1}}{n \sqrt n}=lim_{n\to\infty}\frac{1}{n}(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots+\sqrt{\frac{n-1}{n}})$

$=\int_0^1 \sqrt{x} dx=\int_0^1 x^{\frac{1}{2}} dx=\frac{x^{1+\frac{1}{2}}}{1+\frac{1}{2}} \bigg|_0^1=\frac{x^\frac{2+1}{2}}{\frac{2+1}{2}} \bigg|_0^1=\frac{x^\frac{3}{2}}{\frac{3}{2}}\bigg|_0^1=\frac{2}{3}x^{\frac{3}{2}}\bigg|_0^1 =\frac{2}{3}$

Therefore, $a+b=2+3=\boxed{5}$

A calculus problem by Sutirtha Datta

The answer is 5.

4 solutions

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