Maximillian Patterson

2022-05-01

Prove that $\underset{n\Rightarrow \mathrm{\infty}}{lim}\frac{n}{2n-1}=\frac{1}{2}$

Patricia Duffy

Beginner2022-05-02Added 16 answers

Let $S}_{n$ be a sequence of real numbers, which is said to be convergent to limit L as n approaches to $\mathrm{\infty}$ if

$\mathrm{\forall}\in >0\mathrm{\exists}N\in I$ such that $|{S}_{n}-L|<\in ,\mathrm{\forall}n\ge N$

Given$\underset{n\Rightarrow \mathrm{\infty}}{lim}\frac{n}{2n-1}=\frac{1}{2}$

Here,$S}_{n}=\frac{n}{2n-1},L=\frac{1}{2$

Thus

$|{S}_{n}-L|<\in$

$|\frac{n}{2n-1}-\frac{1}{2}|<\in$

$\left|\frac{2n-1(2n-1)}{2(2n-1)}\right|<\in$

$\left|\frac{2n-2n+1}{2(2n-1)}\right|<\in$

$\left|\frac{1}{4n-2}\right|<\in$

$\frac{1}{4n-2}<\in$

$\frac{1}{4n}<\in$

$4n>\frac{1}{\in}$

$n>\frac{1}{4\in}=N$

Select$N=\left[\frac{1}{4\in}\right]$

$\mathrm{\forall}\in >0\mathrm{\exists}N=\left[\frac{1}{4\in}\right]$ such that $|\frac{n}{2n-1}-\frac{1}{2}|<\in ,\mathrm{\forall}n\ge N$

Hence,$\underset{n\Rightarrow \mathrm{\infty}}{lim}\frac{n}{2n-1}=\frac{1}{2}$

Given

Here,

Thus

Select

Hence,

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