4
$\begingroup$

What is the limit of the sequence $\sin^2 (\pi\sqrt{n^2+n})$ as $n$ tends to infinity?

My Attemp: I replace the square root with $n+\frac 12$ (its equivalent) and the rest is routine:

$\lim \sin^2 (\pi\sqrt{n^2+n}) = \lim \sin^2 (\pi n + \frac{\pi}2) = 1$

Is this correct?

  • 2
    The idea and the result are correct, but you should formulate it a bit more accurate : $$\sqrt{n^2+n}=n\sqrt{1+\frac{1}{n}}=n(1+\frac{1}{2n}+O(\frac{1}{n^2}))=n+\frac{1}{2}+O(\frac{1}{n})$$ – 2017-02-20
  • 0
    @AnotherTest Why? Would you explaine more? – 2017-02-20

2 Answers 2

3

Hint. The sequence is convergent.

As $n$ tends to $+\infty$, we may write $$ \begin{align} u_n &:=\sin^2 \left( \pi \sqrt{n^2+n }\right)\\ &=\sin^2 \left( \pi n \:\sqrt{1+\frac{1}{n}}\right)\\ &=\sin^2 \left( \pi n \:\left(1+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\ &=\sin^2 \left( \pi n +\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\\ &=\left((-1)^n\sin \left(\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\right)^2 \end{align} $$ Can you finish it?

0

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sin^{2}\pars{\pi\root{n^{2} + n}} & = \sin^{2}\pars{\pi n\root{1 + {1 \over n}}} = \sin^{2}\pars{\pi n + \pi n\bracks{\root{1 + {1 \over n}} - 1}} \\[5mm] & = \bracks{\pars{-1}^{n}\sin\pars{\pi n\bracks{\root{1 + {1 \over n}} - 1}}}^{2} \\[5mm] & = \sin^{2}\pars{\pi \over \root{1 + 1/n} + 1} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \sin^{2}\pars{\pi \over 2} = \bbx{\ds{1}} \end{align}