Answer to Question #191550 in Real Analysis for Amara

Question

Find the following limits

lim( sqrt2n+1− Sqrt2n)

n→∞

lim3n^3 −n+8/(4n(n−1)(n−2))

n→∞

b) Consider the sequence (an) = (−1)^n − 2n

 Explain whether the sequence is monotone increasing or decreasing, whether it is monotone and if limn→∞(an) = −∞

Answer 1

(a) (i)$lim_{n\to \infty} (\sqrt{2n+1}-\sqrt{2n})$

          Rationalising the given expression-

                

        $= lim_{n\to \infty} (\sqrt{2n+1}-\sqrt{2n})\times \dfrac{ \sqrt{2n+1}+\sqrt{2n}}{\sqrt{2n+1}+\sqrt{2n}} \\[9pt] = lim_{n\to \infty} \dfrac{ (\sqrt{2n+1})^2-(\sqrt{2n})^2}{\sqrt{2n+1}+\sqrt{2n}} \\[9pt] =lim _{n\to \infty} \dfrac{1}{\sqrt{2n+1}+\sqrt{2n}}$

        Applying limits

         = 0

(ii) $lim_{n\to \infty} \dfrac{3n^3-n+8}{4n(n-1)(n-2)}$

 

    Now writing it my simplifying-

     

   $=lim_{n\to \infty} \dfrac{3n^3-n+8}{4n(n^2-2n-n+2)} \\[9pt] =lim_{n\to \infty} \dfrac{3n^3-n+8}{4n^3-12n^2+8n} \\[9pt]=lim_{n\to \infty} \dfrac{n^3(3-\frac{1}{n^2}+\frac{8}{n^3})}{n^3(4-\frac{12}{n}+\frac{8}{n^2})} \\[9pt]=\dfrac{3-0+0}{4-0+0}=\dfrac{3}{4}$

    

(b) Given, sequence

        $a_n=(-1)^n-2n$

        $a_1=-1-2=-3\\ a_2=1-4=-3\\ a_3=-1-6=-7\\ a_4=-7\\ a_5=-1-10=-11\\ a_6=1-12=-11\\$

$\Rightarrow <a_n>=-3,-3,-7,-7,-11,-11...$

 Here, $a_1\ge a_2\ge a_3\ge a_4\ge a_5\ge a_6...$

we know a sequence $<a_n>$ is monotonic decreasing if $a_{n+1}\le a_n \forall n\in N$

Hence Given sequence is monotone decreasing.

So $lim_{n\to \infty}(a_n)=-\infty.$