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Prove that $\dfrac{1+n}{n^2}$ converges as $n \to \infty$

How do I go about constructing this proof? Can I use the definition that $\operatorname{abs}(a_n - L < \epsilon)$?

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Can you use the squeeze theorem? If so, note that $|x_n|\leq 2n/n^2=2/n$ for all $n\geq 1$. Well, actually, using limit laws is even easier in this case... –  1015 Mar 17 '13 at 15:44
    
I presume where you wrote $\operatorname{abs}(a_n-L<\epsilon)$ you meant $\operatorname{abs}(a_n-L)<\epsilon$. –  Michael Hardy Mar 17 '13 at 16:14
    
@MichaelHardy yep sorry I meant abs(an−L)<ϵ. –  Anona anon Mar 17 '13 at 16:43

2 Answers 2

up vote 2 down vote accepted

$$ \frac{1 +n}{n^2} = \frac 1{n^2} + \frac 1 n = \left ( \frac 1 n \right )^2 + \frac 1 n $$ Now use limit-laws.

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if I had square roots involved in my expression, how would I go about rearranging the expression to find whether the sequence converges or not? thank you –  Anona anon Mar 17 '13 at 19:09
    
In general we have the following: If $f:\mathbb R \rightarrow \mathbb R$ is continuous and the sequence $(a_n)_{n=0}^\infty$ converges to say $a \in \mathbb R$ then $\lim_{n \rightarrow \infty} f(a_n) = f(a)$. Taking suqare roots is continuous. –  André Mar 17 '13 at 22:27
    
It is also easy proved. $f$ is continuous in $a$. So let $\epsilon > 0$. Find $\delta > 0$ such that if $|x-a| < \delta$ then $|f(x)-f(a)| < \epsilon$. Find $N \in \mathbb N $ such that $\forall n \geq N: |a_n-a| < \delta$ which implies $|f(a_n)-f(a)| <\epsilon$ for all $n \geq N$ and thus $(f(a_n))_{n=0}^\infty$ converges to $f(a)$. –  André Mar 17 '13 at 22:35

All you have to know is

  1. $\displaystyle\frac1n$ and $\displaystyle\frac1{n^2}\to 0$ as $n\to\infty$
  2. Arithmetical operations such as addition, division, are continuous, meaning theybehave friendly with limit.

So, we have $$\frac{1+n}{n^2}=\frac1{n^2}+\frac1n\to 0+0=0\,.$$

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