Limit of a series

$\lim_{n \to \infty} \dfrac{(1 + 3 + 5 + ... n~terms) - 2(1 + 2 + 3 + ... n~terms)}{\sqrt{n^2 + 1} + \sqrt{4n^2 - 1}}$

$\lim_{n \to \infty} \dfrac{n^2 - n(n + 1)}{\sqrt{n^2 + 1} + \sqrt{4n^2 - 1}}$

$\lim_{n \to \infty} \dfrac{-n}{n \sqrt{1 + \dfrac{1}{n^2}} + n \sqrt{4 - \dfrac{1}{n^2}}}$

$\lim_{\frac{1}{n} \to 0} \dfrac{-1}{\sqrt{1 + \dfrac{1}{n^2}} + \sqrt{4 - \dfrac{1}{n^2}}}$

$\implies - \dfrac{1}{\sqrt{1} + \sqrt{4}} = - \dfrac{1}{1 + 2} = - \dfrac{1}{3} = - \dfrac{a}{b}$

$\implies a \times b = 3$

$\begin{aligned} L & = \lim_{n \to \infty} \frac {1-2+3-4+5-6+\cdots -2n}{\sqrt{n^2+1}+\sqrt{4n^2-1}} \\ & = \lim_{n \to \infty} \frac {(1+2+3+\cdots + 2n) - 4(1+2+3+\cdots+n)}{\sqrt{n^2+1}+\sqrt{4n^2-1}} \\ & = \lim_{n \to \infty} \frac {\frac {2n(2n+1)}2 - 4\left(\frac {n(n+1)}2\right)}{\sqrt{n^2+1}+\sqrt{4n^2-1}} \\ & = \lim_{n \to \infty} \frac {-n}{\sqrt{n^2+1}+\sqrt{4n^2-1}} \\ & = \lim_{n \to \infty} \frac {-n}{n\sqrt{1+\frac 1{n^2}}+2n\sqrt{1-\frac 1{4n^2}}} & \small \color{#3D99F6} \text{Divide up and down by }n \\ & = \lim_{n \to \infty} \frac {-1}{\sqrt{1+\frac 1{n^2}}+2\sqrt{1-\frac 1{4n^2}}} \\ & = \frac {-1}{1+2} = - \frac 13 \end{aligned}$

Therefore, $ab = 1\times 3 = \boxed 3$ .