Roots and Riemann

Calculus Level 3

$\large \lim_{n \to\ \infty} \left(\frac{1}{\sqrt{n}\sqrt{n + 1}} + \frac{1}{\sqrt{n}\sqrt{n + 2}} + \cdots + \frac{1}{\sqrt{n}\sqrt{n + n}} \right)$

The value of the limit above can be expressed as $\displaystyle a\sqrt{b} - \frac cd$ , where $a$ , $b$ , $c$ , and $d$ are positive integers, $\gcd(c, d) = 1$ , and $b$ is square-free.

Find $a^4 + b^3 + c^2 + d$ .

The answer is 29.

1 solution

Zach Abueg
May 25, 2017

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{n \to\ \infty} \Bigg(\frac{1}{\sqrt{n}\sqrt{n + 1}} + \frac{1}{\sqrt{n}\sqrt{n + 2}} + \ \cdots \ + \ \frac{1}{\sqrt{n}\sqrt{n + n}} \Bigg) \\ & = \lim_{n \to\ \infty} \ \sum_{i \ = \ 1}^{n} \frac {1}{\sqrt{n} \sqrt{n + i}} & \small \color{#3D99F6} \text{Rewrite as a Riemann sum: }\lim_{n \to\ \infty} \sum_{i \ = \ 1}^{n} \frac 1n \cdot f\bigg(\frac in\bigg) \\ & = \lim_{n \to\ \infty} \ \sum_{i \ = \ 1}^{n} \ \Bigg(\frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{n\big(1 + \frac in\big)}}\Bigg) \\ & = \lim_{n \to\ \infty} \ \sum_{i \ = \ 1}^{n} \ \Bigg(\frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{n}} \cdot \frac {1}{\sqrt{1 + \frac in}}\Bigg) \\ & = \lim_{n \to\ \infty} \ \sum_{i \ = \ 1}^{n} \ \Bigg(\frac 1n \cdot \frac {1}{\sqrt{1 + \frac in}}\Bigg) & \small \color{#3D99F6} \lim_{n \to\ \infty} \sum_{i \ = \ 1}^{n} \frac 1n \cdot f\bigg(\frac in\bigg) = \int_0^1 f(x) \ dx \\ & = \int_0^1 \frac{1}{\sqrt{1 + x}} \ dx \\ & = 2\sqrt{1 + x} \ \bigg|_0^1 \\ & = 2\sqrt{2} - 2 \end{aligned}$

$\displaystyle \implies a^4 + b^3 + c^2 + d = 2^4 + 2^3 + 2^2 + 1 = 29$

Roots and Riemann

The answer is 29.

1 solution

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