Question

L $\lim_{n \to \infty} \frac{1}{\sqrt{n^2-1}} + \frac{1}{\sqrt{n^2-4}} + \frac{1}{\sqrt{n^2-9}} + \dots + \frac{1}{\sqrt{2n-1}}$

Answer 1

$\frac{\pi}{2}$

Answer 2

The given sum can be written as $S_n = \sum_{k=1}^{n-1} \frac{1}{\sqrt{n^2-k^2}}$. We can rewrite the general term by factoring out $n^2$ from the square root: $\frac{1}{\sqrt{n^2-k^2}} = \frac{1}{\sqrt{n^2\left(1 - \frac{k^2}{n^2}\right)}} = \frac{1}{n\sqrt{1 - \left(\frac{k}{n}\right)^2}}$ So, the sum becomes: $S_n = \sum_{k=1}^{n-1} \frac{1}{n} \cdot \frac{1}{\sqrt{1 - \left(\frac{k}{n}\right)^2}}$ This is a Riemann sum for the function $f(x) = \frac{1}{\sqrt{1-x^2}}$ over the interval $[0, 1]$. As $n \to \infty$, the limit of this sum is the definite integral: $L = \lim_{n \to \infty} S_n = \int_0^1 \frac{1}{\sqrt{1-x^2}} dx$ The integral of $\frac{1}{\sqrt{1-x^2}}$ is $\arcsin(x)$. Evaluating the definite integral: $\int_0^1 \frac{1}{\sqrt{1-x^2}} dx = [\arcsin(x)]_0^1 = \arcsin(1) - \arcsin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ Therefore, the limit is $\frac{\pi}{2}$.