Exercise 2.2.2

Question

Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.
  \begin{enumerate}[label=(\alph*)] 
  \item $\lim \frac{2 n+1}{5 n+4}=\frac{2}{5}$.
  \item $\lim \frac{2 n^{2}}{n^{3}+3}=0$.
  \item $\lim \frac{\sin \left(n^{2}\right)}{\sqrt[3]{n}}=0$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item We have $$ \left|\frac{2n + 1}{5n + 4} - \frac{2}{5}\right| = \left|\frac{5(2n + 1) - 2(5n + 4)}{5(5n + 4)}\right| = \left|\frac{-3}{5(5n + 4)}\right| = \frac{3}{5(5n + 4)} < \epsilon $$ We now find $n$ such that the distance is less then $\epsilon$ $$ \frac{3}{5(5n + 4)} < \frac{1}{n} < \epsilon \implies n > \frac{1}{\epsilon} $$ You could also solve for the smallest $n$, which would give you $$ \frac{3}{5(5n + 4)} < \epsilon \implies 5n + 4 > \frac{3}{5\epsilon} \implies n > \frac{3}{25\epsilon} - \frac{4}{5} $$ I prefer the first approach, the second is better if you were doing numerical analysis and wanted a precise convergence rate. \item We have $$ \left|\frac{2n^2}{n^3 + 3} - 0\right| = \frac{2n^2}{n^3 + 3} < \frac{2n^2}{n^3} = \frac{2}{n} < \epsilon \implies n > \frac{2}{\epsilon} $$ \item We have $$ \frac{\sin(n^2)}{n^{1/3}} \le \frac{1}{n^{1/3}} < \epsilon \implies n > \frac{1}{\epsilon^3} $$ Really slow convergence! if $\epsilon = 10^{-2}$ we would require $n > 10^6$ \end{enumerate}

Exercise 2.2.2

Answers

Comments

Exercise 2.2.2

Answers

Comments

Add answer