If we accept the fact that the sequence nn + 1+n=1converges to the limit L = 1, then

Chapter 9, Problem 49

(choose chapter or problem)

If we accept the fact that the sequence

$\left\{\frac{n}{n+1}\right\}_{n=1}^{+\infty}$

converges to the limit $L=1$, then according to Definition 9.1.2, for every $\epsilon>0$ there exists an integer $N$ such that

$\left|a_{n}-L\right|=\left|\frac{n}{n+1}-1\right|<\epsilon$

when $n \geq N$ . In each part, find the smallest value of $N$ for the given value of $\epsilon$

(a) $\epsilon=0.25$ (b) $\epsilon=0.1$ (c) $\epsilon=0.001$

Equation Transcription:

${\left\{{\frac{n}{n+1}}\right\}}_{n=1}^{+\infty{}}$

$L=1$

$\epsilon{}>0$

$N$

$\left|{{a}_{n}-L}\right|=\left|{\frac{n}{n+1}-1}\right|<\epsilon{}$

$n\geq{}N$

$\epsilon{}$

$\epsilon{}=0.25$

$\epsilon{}=0.1$

$\epsilon{}=0.001$

Text Transcription:

{n/n+1}_n=1 ^+infinity

L=1

epsilon > 0

|a_n - L| = |n/n+1 - 1| < epsilon

n geq N

epsilon

epsilon = 0.25

epsilon = 0.1

epsilon = 0.001

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