Chant6j

2022-07-02

I have to evaluate
$\underset{n\to \mathrm{\infty }}{lim}\left(-1{\right)}^{n-1}\mathrm{sin}\left(\pi \sqrt{{n}^{2}+0.5n+1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}ϵ\phantom{\rule{thinmathspace}{0ex}}N$

Maggie Bowman

Expert

The problem is that as $n\to \mathrm{\infty }$, $\sqrt{{n}^{2}+0.5n+1}$ diverges, and $\sqrt{{n}^{2}+0.5n+1}-n\ne 0$
. We have
$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\sqrt{{n}^{2}+0.5n+1}-n& =\underset{n\to \mathrm{\infty }}{lim}\sqrt{\left(n+0.25{\right)}^{2}+1-{0.25}^{2}}-n\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{\left(\sqrt{{n}^{2}+0.5n+1}-n\right)\left(\sqrt{{n}^{2}+0.5n+1}+n\right)}{\sqrt{{n}^{2}+0.5n+1}+n}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{0.5n+1}{\sqrt{{n}^{2}+0.5n+1}+n}=\frac{1}{4}\end{array}$

Esmeralda Lane

Expert

Before $1+\frac{0.5}{n}+\frac{1}{{n}^{2}}$ can "become" 1, it is greater than 1. And so when it multiplies by ${n}^{2}$, you get something greater than ${n}^{2}$. And it turns out that it is still greater enough than ${n}^{2}$ to make a difference.
Use ${n}^{2}+0.5n+1=\left(n+0.25{\right)}^{2}+\frac{15}{16}$. Then you have $\left(n+0.25\right)\sqrt{1+\frac{15}{16\left(n+0.25{\right)}^{2}}}$. At this point your intuition works out to be correct, and $1+\frac{15}{16\left(n+0.25{\right)}^{2}}$ goes to 1 quickly enough. But more should be done to prove this.

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