Chant6j

Answered

2022-07-02

I have to evaluate

$\underset{n\to \mathrm{\infty}}{lim}(-1{)}^{n-1}\mathrm{sin}(\pi \sqrt{{n}^{2}+0.5n+1})\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}\u03f5\phantom{\rule{thinmathspace}{0ex}}N$

$\underset{n\to \mathrm{\infty}}{lim}(-1{)}^{n-1}\mathrm{sin}(\pi \sqrt{{n}^{2}+0.5n+1})\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}\u03f5\phantom{\rule{thinmathspace}{0ex}}N$

Answer & Explanation

Maggie Bowman

Expert

2022-07-03Added 14 answers

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

$\begin{array}{rl}\underset{n\to \mathrm{\infty}}{lim}\sqrt{{n}^{2}+0.5n+1}-n& =\underset{n\to \mathrm{\infty}}{lim}\sqrt{(n+0.25{)}^{2}+1-{0.25}^{2}}-n\\ & =\underset{n\to \mathrm{\infty}}{lim}\frac{(\sqrt{{n}^{2}+0.5n+1}-n)(\sqrt{{n}^{2}+0.5n+1}+n)}{\sqrt{{n}^{2}+0.5n+1}+n}\\ & =\underset{n\to \mathrm{\infty}}{lim}\frac{0.5n+1}{\sqrt{{n}^{2}+0.5n+1}+n}=\frac{1}{4}\end{array}$

. We have

$\begin{array}{rl}\underset{n\to \mathrm{\infty}}{lim}\sqrt{{n}^{2}+0.5n+1}-n& =\underset{n\to \mathrm{\infty}}{lim}\sqrt{(n+0.25{)}^{2}+1-{0.25}^{2}}-n\\ & =\underset{n\to \mathrm{\infty}}{lim}\frac{(\sqrt{{n}^{2}+0.5n+1}-n)(\sqrt{{n}^{2}+0.5n+1}+n)}{\sqrt{{n}^{2}+0.5n+1}+n}\\ & =\underset{n\to \mathrm{\infty}}{lim}\frac{0.5n+1}{\sqrt{{n}^{2}+0.5n+1}+n}=\frac{1}{4}\end{array}$

Esmeralda Lane

Expert

2022-07-04Added 7 answers

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 ${n}^{2}+0.5n+1=(n+0.25{)}^{2}+\frac{15}{16}$. Then you have $(n+0.25)\sqrt{1+\frac{15}{16(n+0.25{)}^{2}}}$. At this point your intuition works out to be correct, and $1+\frac{15}{16(n+0.25{)}^{2}}$ goes to 1 quickly enough. But more should be done to prove this.

Use ${n}^{2}+0.5n+1=(n+0.25{)}^{2}+\frac{15}{16}$. Then you have $(n+0.25)\sqrt{1+\frac{15}{16(n+0.25{)}^{2}}}$. At this point your intuition works out to be correct, and $1+\frac{15}{16(n+0.25{)}^{2}}$ goes to 1 quickly enough. But more should be done to prove this.

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