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$\dfrac{3n}{n+4}$ where $n$ is an element of the natural numbers.

I used calculus to find my supposed surprema and infimum. But I am having a difficult time proving they are infimum and suprema. I am assuming to prove the infimum and suprema you have to make a inequality about the sequence, i.e.

$\dfrac{3n}{n+4} + 3 > \epsilon$

But I am having a hard time understanding also the infimum

$\epsilon < \dfrac{3n}{n+4} - \dfrac{3}{5}$

Please help

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    Try proving by contradiction. Suppose that there is an n such that $3/5 > 3n/(n+4)$ and the same thing for the upper bound. – 2017-01-31
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    HINT: $$\frac{3n}{n+4} = 3 - \frac{12}{n+4}$$ – 2017-01-31
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    Who said to find supremum and infimum? If someone says "bounded above by $3$," all they mean is that the $n$th term is never greater than $3$. Likewise, bounded below by $3/5$ means only the $n$th term is always greater than $3/5.$ No epsilons need apply. – 2017-01-31
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    the question in my assignment wants me to prove it's a infimum or suprema – 2017-01-31

1 Answers 1

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$$\frac{n}{n+4}=\frac{n+4-4}{n+4}=1-\frac{4}{n+4}$$ Now, as $n$ increases, $n+4$ also increases, so $\frac{1}{n+4}$ decreases, and $\frac{-4}{n+4}$ increases.

However, $\frac{-4}{n+4}$ is negative for all $n\ge 0$, so it is increasing and negative, with limit $0$. Hence by taking $n$ large enough, you can make $\frac{-4}{n+4}$ as close to zero as you want. Now do some algebra to see how big $n$ has to be to make $\frac{3n}{n+4}>\frac{3}{5}$.

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    i am not sure how this proves it's an infimum or suprema – 2017-01-31