Prove that $a_n=\frac{n}{n+1}$ converges

Question

We know a sequence is convergent if and only if given $\epsilon>0$ there exist $n_0\in \mathbb{N}$(which depends on $\epsilon>0$) such that $\forall n\ge n_0, \lvert a_n - L\rvert \lt \epsilon$.

I already know $\displaystyle \lim_{n \to \infty}\frac{n}{n+1}=1$

Then, using the definition we want to find that $\epsilon>0$ such that: $$\left\lvert \frac{n}{n+1}-1\right\rvert\lt \epsilon$$ We can rewrite this as : $$\left\lvert \frac{-1}{n+1}\right\rvert\lt \epsilon$$

The problem comes here , where I do not really know how to proceed and how to get rid of the absolute value, any help would be appreciated.

Answer 1

$$\left \lvert \frac{-1}{n+1} \right\rvert = \frac{1}{n+1} \lt \epsilon \iff n > \frac1\epsilon - 1$$ So take $n_0 = \left \lfloor \dfrac1\epsilon \right\rfloor$ to guarantee that for all $n \ge n_0$, the above inequality holds.

Answer 2

First, fix $\varepsilon>0$. Then you have to find $n_0=n_0(\varepsilon)$ such that $$ \left|\frac{-1}{n+1}\right|<\varepsilon $$ for all $n\ge n_0$. Just choose $n_0=\lceil 1/\varepsilon \rceil$.

Answer 3

I think the issue is more one of manipulating absolute values. The absolute value of a product (quotient) is the product (quotient) of the absolute values, so $$ \left|\frac{-1}{n+1} \right|=\frac{|-1|}{|n+1|}=\frac{1}{n+1} $$ since the absolute value of a negative number is its negative, and the absolute value of a positive number, which $n+1$ is, is itself.

Answer 4

$$ a_n=\frac{1}{1+1/n} $$ Then $$ \left|a_n-1\right|=\left|\frac{1/n}{1+1/n}\right|=\left|1-\frac{1}{n}\right| $$ And it is clear that $$ \left|1-\left(1-\frac{1}{n}\right)\right| \underset{n\rightarrow +\infty}{\rightarrow}0 $$

Answer 5

$$|\frac {-1}{n+1}|=\frac {1}{n+1}<\frac {1}{n} $$

It is sufficient to find $n_0$ such that

$$n\ge n_0\implies \frac {1}{n}<\epsilon $$ or $$n\ge n_0\implies n>\frac {1}{\epsilon} $$ One can take $$n_0=\lfloor \frac {1}{\epsilon}\rfloor +2019$$

Answer 6

Given $\epsilon.$

Archimedes:

There is a $n_0 +1 > 1/\epsilon$, $n_0 \in \mathbb{N}$

For $n > n_0 $:

$\dfrac{1}{n+1} \lt \dfrac{1}{n_0+1} \lt \epsilon.$

Answer 7

Hint: What is $1 - \frac n{n+1}$?

It is $1 - \frac n{n+1} = \frac {n+1}n -\frac n{n+1} = \frac 1{n+1} > 0$.

So $|1 - \frac n{n+1}| = |\frac 1{n+1}| = \frac 1{n+1}$

When can we say that $\frac 1{n+1} < \epsilon$?

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Well, we can say that whenever $n+1 > \frac 1{\epsilon}$.

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So, for any $\epsilon > 0$, let $n_0 \ge \frac 1{\epsilon}$.

then $n \ge n_0$ would mean $n+1 > \frac 1{\epsilon}$ and $\epsilon > \frac 1{n+1} = 1-\frac n{n+1} = |1 - \frac n{n+1}|$.

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"The problem comes here , where I do not really know how to proceed and how to get rid of the absolute value"

Notice that all of the following statements are true.

$n > 0$ and $n + 1 > 0$ and $-1 < 0$ and $\frac {-1}{n+1} < 0$ and $1 > 0$ and $\frac {1}{n+1} > 0$ and $n < n+1$ and $\frac n{n+1} < 1$ and $1 - \frac n{n+1} > 0$, etc. ad nauseum.

And that's that.