The value of _n1+2-3+4+5-6++(3n-2)+(3n-1)-3n√2n^4+4n+3-n^4+5n+4 is:

Question

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

Accepted Answer

Each triple $(3k-2)+(3k-1)-3k = 3k-3$, so the numerator's running total (grouped in triples) equals $0+3+6+\cdots$ ($n$ terms) $= \dfrac{3n(n-1)}2$.

The denominator behaves like $n^2\sqrt2-n^2 = n^2(\sqrt2-1)$ for large $n$.

$$\lim_{n\to\infty}\frac{3n(n-1)/2}{n^2(\sqrt2-1)} = \frac{3/2}{\sqrt2-1} = \frac32(\sqrt2+1)$$

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