Question

The value of $\lim_{n \to \infty} \frac{1+2-3+4+5-6+.......+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$ is:

• A. 3($\sqrt{2}$+1)
• B. $\frac{3}{2\sqrt{2}}$
• C. $\frac{3}{2}(\sqrt{2}+1)$
• D. $\frac{\sqrt{2}+1}{2}$

Answer 1

Answer: (C)

$\frac{3}{2}(\sqrt{2}+1)$

Answer 2

The numerator series can be grouped as $(1+2-3) + (4+5-6) + \dots + ((3n-2)+(3n-1)-3n)$. Each group sums to $0, 3, 6, \dots, 3n-3$. The sum of these $n$ terms is an arithmetic progression: $\frac{n}{2}(0 + 3n-3) = \frac{3n(n-1)}{2}$.

The denominator is $D_n = \sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}$. Rationalizing the denominator gives: $D_n = \frac{(2n^4+4n+3)-(n^4+5n+4)}{\sqrt{2n^4+4n+3}+\sqrt{n^4+5n+4}} = \frac{n^4-n-1}{\sqrt{2n^4+4n+3}+\sqrt{n^4+5n+4}}$.

For large $n$, the dominant terms are: Numerator $\sim \frac{3n^2}{2}$. Denominator $\sim \frac{n^4}{\sqrt{2n^4}+\sqrt{n^4}} = \frac{n^4}{\sqrt{2}n^2+n^2} = \frac{n^4}{(\sqrt{2}+1)n^2} = \frac{n^2}{\sqrt{2}+1}$.

The limit is: $\lim_{n \to \infty} \frac{\frac{3n^2}{2}}{\frac{n^2}{\sqrt{2}+1}} = \frac{3}{2}(\sqrt{2}+1)$