The value of _ n → ∞ 1 + 2 - 3 + 4 + 5 - 6 + + ( 3n - 2 ) + ( 3n -...

The given limit $\lim_{n \to \infty}\dfrac{1 + 2 - 3 + 4 + 5 - 6 + \ldots + ( 3n - 2 ) + ( 3n - 1 ) - 3n}{\sqrt{2{n}^{4} + 4n + 3}-\sqrt{{n}^{4} + 5n + 4}}$ can be expressed as,$\lim_{n \to \infty}\dfrac{\mathop{∑}_{r = 1}^{n}\left(3r - 2\right) + \left(3r - 1\right) - 3r}{\sqrt{2{n}^{4} + 4n + 3} - \sqrt{{n}^{4} + 5n + 4}}$

$= \lim_{n \to \infty}\dfrac{\mathop{∑}_{r = 1}^{n}3\left(r - 1\right)}{\sqrt{2{n}^{4} + 4n + 3} - \sqrt{{n}^{4} + 5n + 4}}$

$= \lim_{n \to \infty}\dfrac{0 + 3 + 6 + 9 + \ldots . nterms}{\sqrt{2{n}^{4} + 4n + 3} - \sqrt{{n}^{4} + 5n + 4}} . . . . . . . ( 1 )$

We know the sum of the nth term of A.P. is given by

${S}_{n} = \dfrac{n}{2}\left[2a + \left(n - 1\right)d\right]$

Using above formula in equation(1) we get,

$= \lim_{n \to \infty}\dfrac{\dfrac{3n ( n - 1 )}{2}}{\left(\sqrt{2{n}^{4} + 4n + 3} - \sqrt{{n}^{4} + 5n + 4}\right)}$

$= \lim_{n \to \infty}\dfrac{3n ( n - 1 )}{2\left(\sqrt{2{n}^{4} + 4n + 3} - \sqrt{{n}^{4} + 5n + 4}\right)}$

$= \lim_{n \to \infty}\dfrac{3 ( 1 - \dfrac{1}{n} )}{2\left(\sqrt{2 + \dfrac{4}{{n}^{3}} + \dfrac{3}{{n}^{4}}} - \sqrt{1 + \dfrac{5}{{n}^{3}} + \dfrac{4}{{n}^{4}}}\right)}$

$= \dfrac{3 ( 1 - 0 )}{2\left(\sqrt{2 + 0 + 0} - \sqrt{1 + 0 + 0}\right)}$

$= \dfrac{3}{2 ( \sqrt{2} - 1 )} = \dfrac{3}{2} ( \sqrt{2} + 1 )$