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Please help me check limit of the sequence $a_n=\frac{n}{n+1}$

  • 2
    Hint: $a_n = 1-\frac1{n+1}$.2012-09-18

4 Answers 4

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It is a consequence of the unboundedness of the natural numbers that, for any $\epsilon >0$, there exists a natural number $n$ for which $\dfrac 1 n <\epsilon$. This is used to prove the "basic" limit,

$\tag 1 \lim\limits_{n\to \infty}\frac 1 n = 0$

The expression $\dfrac n {n+1}$ can be written as

$1-\frac 1 {n+1}$

The algebra of limits says that if $\lim\limits_{n\to \infty}a_n=a$ and $\lim\limits_{n\to \infty} b_n=b$ , then $\lim\limits_{n\to \infty} c_n = a+b$ where $c_n=a_n+b_n$. It is straightforward that $\lim\limits_{n\to \infty} 1 =1$, so you need to show

$\tag 2 \lim_{n\to \infty}\frac {-1} {n+1} $

exists and equals something. Can you do this using $(1)$? Hint: if $n$ is a natural number, so is $n+1$, and $\left|-\frac 1 {n+1}-0\right|=\frac 1 {n+1}$

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HINT: If you look at the numbers $\frac{n}{n+1}$ for large $n$, you should be able to make a good guess. To prove your guess, note that $\frac{n}{n+1}=1-\frac1{n+1}\;.$

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

$1 for $n\in N$ $\Rightarrow$ $\frac{1}{n+1}<1$ $\Rightarrow$ $1-\frac{1}{n+1}<1$.

Definitely

$|a_n|<1$ $\Rightarrow$ $-1

1

Consider the function $f(x)=\frac{x}{x+1}$. We have $\lim_{x \to \infty} \frac{x}{x+1}=\lim_{x \to \infty} \frac {1}{1+(1/x)}$. Do you see how to finish?

  • 0
    Curious. Note, however, that a sequence $f(n)$ might have a limit while the associated function $f(x)$ might fail to have a limit for $x\to \infty$. The converse is not true: if $\lim f(x)=L$ for $x\to \infty$, then necessarily $\lim f(n)=L$.2012-09-18