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I'm given the following:

$(a_n)_{n\in\mathbb{N}} $ is a real sequence defined as $a_n := 1/\sqrt{n} $ for $n \in \mathbb{N}$.

i) Show that $(a_n)_{n\in\mathbb{N}} $ is bounded from above and below, and give the explicit boundaries.

ii) Show that, $(a_n)_{n\in\mathbb{N}} $, is monotonely decreasing.

iii) Give $\lim_{n\to\infty} a_n $

Now I understand all of this, Im just not sure how to write a good mathematical proof for it.

Basically my idea is like this... Since n can only be a positive integer, when n = 1, it will basically be $1/\sqrt{1} = 1$. When n > 1, then the fraction will become smaller, and thats why the above boundary is 1. As n approaches inifinity, then the fraction $1/\sqrt{n}$ will approach 0. So the lower boundary is 0.

The function is monotone decreasing since ($1/\sqrt{n}) \geq (1/\sqrt{n+1})$.

And $\lim_{n\to\infty} a_n = 0 $

This is all clear to me, but I just feel like if I write it like this, it is not really a mathematical proof and I am not sure how I can formulate this as a proper valid proof.

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    What you ask for is called sequence and not series. – 2017-01-29
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    What you wrote is clear to me. as a techer I'd accept it as is. Your instructor might want more formality, perhaps an algebraic proof of the inequality you state. – 2017-01-29
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    Yeah because I remember that one one previous paper, I wrote whatever made sense to me, but it wasnt really mathematicallz structured, and I got 0 points for it... Although it was correct... – 2017-01-29

2 Answers 2

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I think you may want to clean up your answer to (i) a bit, though it is already correct. Here's one possible way to write it:

(i) Let $n \geq 1$. It follows that $a_n$ is trivially positive, and hence $0$ is a lower bound of the sequence $(a_n)_n$. Moreover, $\sqrt{n} \geq 1$ and hence $a_n \leq 1$. As such, $1$ is an upper bound of the sequence $(a_n)_n$.

You should expand upon your answer to (iii):

Hint: (iii) Since the sequence $(a_n)_n$ is nonincreasing with a lower bound of $0$, it must admit a limit. Call this limit $a$. Taking limits on both sides of the inequality $a_n \geq 0$, we get that $a \geq 0$. Now, suppose $a > 0$, and arrive at a contradiction. Conclude that $a=0$.

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For all $n \in \mathbb{N}\setminus \{ 0\},$ one has $n \geq 1,$ hence $\sqrt{n}\geq \sqrt{1}=1$ and thus $\displaystyle 1 \geq \frac{1}{\sqrt{n}},$ so $a_n$ is bounded above by $1$. We also have (In particular) $n > 0$ for all $n$, hence $\displaystyle \frac{1}{n}>0$ and therefore $\displaystyle \frac{1}{\sqrt{n}}>0$, so the sequence is bounded below by $0$.

Furthermore, for all $n \in \mathbb{N}\setminus \{ 0\}$ we have $n+1 > n,$ hence $\sqrt{n+1}>\sqrt{n}$ and thus $\displaystyle \frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}},$ so the sequence is monotone decreasing.

Finally, given $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\displaystyle \epsilon > \frac{1}{N}$ (Archimedean property), so for all $n \geq N^2$ we have $$ |a_n-0|=\frac{1}{\sqrt{n}} \leq \frac{1}{N} <\epsilon,$$ so $\lim_{n \to \infty}a_n=0.$