MathsGee Answers is Zero-Rated (You do not need data to access) on: Telkom | Dimension Data | Rain | MWEB
First time here? Checkout the FAQs!
x
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
9 views
Consider the real sequence $\left\{x_{n}\right\}$ given by $$ x_{n}=\frac{1}{\ln n} $$ for $n \geq 2$. Provide an $\varepsilon-N$ proof that $x_{n} \rightarrow 0$.
in Mathematics by Bronze Status (8,688 points) | 9 views

1 Answer

0 like 0 dislike
Best answer
Given $\varepsilon>0$, let $N=e^{1 / \varepsilon}$. Then, for all $n>N=e^{1 / \varepsilon}$, we have that
$$
e^{1 / \varepsilon}<n
$$
Taking the natural logarithm of both sides, we have that
$$
\frac{1}{\varepsilon}<\ln n
$$
Since $n \geq 2, \ln n>0$ and $\varepsilon>0 .$ So we can cross divide to get that
$$
\frac{1}{\ln }<\varepsilon
$$
This is equivalent to
$$
\begin{array}{l}
\left|\frac{1}{\ln n}-0\right|<\varepsilon \\
\text { So, } 1 / \ln n \rightarrow 0
\end{array}
$$
by Bronze Status (8,688 points)

Related questions

MathsGee provides answers to subject-specific educational questions for improved outcomes.



On MathsGee Answers, you can:


1. Ask questions
2. Answer questions
3. Comment on Answers
4. Vote on Questions and Answers
5. Donate to your favourite users

MathsGee Tools

Math Worksheet Generator

Math Algebra Solver

Trigonometry Simulations

Vectors Simulations

Matrix Arithmetic Simulations

Matrix Transformations Simulations

Quadratic Equations Simulations

Probability & Statistics Simulations

PHET Simulations

Visual Statistics

ZeroEd Search Engine

Other Tools

MathsGee ZOOM | eBook