Please help me check limit of the sequence $a_n=\frac{n}{n+1}$
Limit of the sequence $a_n=\frac{n}{n+1}$
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2Hint: $a_n = 1-\frac1{n+1}$. – 2012-09-18
4 Answers
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 $\left|-\frac 1 {n+1}-0\right|=\frac 1 {n+1}$
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 $\frac{n}{n+1}=1-\frac1{n+1}\;.$
$|a_n|=|\frac{n}{n+1}|$ $=|\frac{n+1-1}{n+1}|$ $=|\frac{n+1}{n+1}-\frac{1}{n+1}|$ $=|1-\frac{1}{n+1}|$
$1
Definitely
$|a_n|<1$ $\Rightarrow$ $-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?
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0Curious. 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