Prove that the sequence $f(n)$ = $(1/(n)^{0.5}) ,∀n ∈ N $converges towards 0. This is a homework problem for a university math proofs course and I am getting stuck on all convergence/divergence problems currently.
Here is the definition of convergence to a number 'L':
∀$\epsilon$ > 0, ∃ m ∈ N such that ∀n ∈ N, (n ≥ m =⇒ |un − L < $\epsilon$).
For someone who is good at these kinds of limit proofs, what is your thought process when you go about them?
Right now I am first trying to find specific examples of the variables that will show that the definition is true. Or I will negate the definition and try to find specific examples that will show that the negation is false.
But I'm confused if I can do this because when I have statements in the definition such as "for every n", I can't possible guess and check an infinite number of values so what do I do?
Thanks,
John
 A: Here's the catch :

To check that some condition holds for all $n$, you need  not go and test all the $n$ and verify that it holds. That's what we have algebra for : showing something holds true (as an algebraic (in)equality) for a variable means it holds true as a numerical (in)equality for all values that variable can take.

For example, to verify that $(n+1)^2 > 0$ for all natural numbers $n$, you don't need to check that $(1+1)^2 > 0$, $(2+1)^2 > 0$ , $(3+1)^2 > 0$ etc. , but simply observe that if $n$ is a natural number then $n+1 > 0$ and so $(n+1)^2$ is the product of two positive things, hence positive for all natural numbers $n$.
From here, the fact that $(81618+1)^2 > 0$ is proved. We need not go and find what $(81618+1)^2$ is, to check this. (It, in fact, is equal to $6661661161$).  

For the above question, what you need to do, is to "understand" the sequence $u_n$. Maybe plot $u_n$ on an imaginary graph. Look at $u_n$ for large values of $n$.
The idea of "the limit of $u_n$" is : where does this sequence  $u_n$ end? Where do the terms eventually all go towards? Whatever number this $u_n$ appears to be ending at, is a suitable candidate for the limit, namely $L$.
Now, computing $u_n$ for $n$ not being a square is a bit of an issue, but for example, we can evaluate $u_n$ at some large powers of $10$, say.
$$u_{100} = \frac 1{10} , u_{10000} = \frac 1{100} , u_{1000000} = \frac 1{1000} , ...$$ 
Now, it is not very difficult to notice a pattern : the numbers $\frac 1{10}, \frac 1{100}, \frac 1{1000}$ all seem to be approaching zero : if I increase $m$ in $u_m$, the number seems to be going closer to zero. Therefore, one thinks that $0$ ca be a suitable candidate for $L$.
However, one cannot jump to conclusions by just looking at one "sub"-sequence. I took $u_{100}, u_{10000}, u_{1000000}$ above. What if instead, I took say some other arbitrary sequence $u_{3}, u_{45}, u_{765}, u_{45646} , ...$ and found that the values were approaching some point other than zero? 
Since you are a beginner, you have the licence to play around and check what happens to any such subsequence. For example, try out : $u_1,u_4,u_9,...$ or try out $u_{4}, u_{16}, u_{64}, u_{256},...$.
What you will notice, is that all the values are still decreasing to zero.
Taking about three or four such "sub"sequences, you will be able to make a strong intuitive case for zero being the limit of the sequence itself.
However, intuition is (at least in this case) not the sword that mathematics lives and dies by : we need to check that $0$ a limit from the definition of something being a limit of a sequence.

For this, we need to understand what the definition wants to show : Given an $\epsilon$, how to choose an $m$ such that $\forall n \geq m$ , $|u_n - L| < \epsilon$ occurs? 
Think of it this way : $\epsilon$ is given to you like a constraint parameter : you need to ensure that $|u_n - L| < \epsilon$ happens for whatever $\epsilon$ given to you. For this , you need to choose $m$.
Now, imagine you have a house, and some of the usual things in that house. You want to shift some object in, and need to make space for it. If the object is big, then you need a lot of space, and need to rearrange a lot of things. If the object is small, then you need very little space to fit it, so you don't need  to shift too many things.
Key point : without knowing the size of what you want to shift, you don't know how hard you have to work!
Applying that principle here, we want to ultimately have $|u_n - L| < \epsilon$. Unless we decipher this expression, we cannot find the $m$, right?
In general, while solving a limit, always ensure you have completely deciphered the part $|u_n - L| < \epsilon$ first. That will tell you what your final objective is : this helps in finding the $m$.

However, you have a candidate for $L$, namely $L = 0$. You know what $u_n$ is. So, you know what $|u_n - L|$ is :
$$
|u_n - L| = \left|\frac 1{\sqrt n} - 0\right|  = \frac{1}{\sqrt n}
$$
(since $L = 0$ and $u_n$ is a positive quantity, you should not be too surprised that the right side is just $u_n$ itself).
So, the desired expression $|u_n - L| < \epsilon$ is :
$$
\frac 1{\sqrt n} < \epsilon \tag{1}
$$
In the house context, you know what you want to shift in : now to finding $m$.
Recall how to find $m$ : any $m$ such that the inequality $(1)$ holds for every $n \geq m$ will suffice for the purposes.
Now, since $n$ depends upon $m$, we will algebraically rearrange $(1)$ so that we can isolate $n$ separately :
$$
\frac 1{\sqrt n} < \epsilon \iff \frac 1 \epsilon < \sqrt n \iff \frac 1{\epsilon^2} < n \tag{(2)}
$$
Excellent. Now, we have to choose $m$ such that for every $n \geq m$, the inequality $(2)$ holds. Is this difficult?
Well, simply observing $(2)$ gives us a candidate for $m$ : we simply take any natural number $m$ such that $m > \frac 1{\epsilon^2}$. Now, if $n \geq m$, then :$$
n \geq m > \frac 1{\epsilon^2} \implies n > \frac 1{\epsilon^2} \tag{3}
$$
Key point : The way to pick $m$ is to write $n$ as the subject of $|u_n - L| < \epsilon$, as we did in $(2)$. Then, with a little work the desired $m$ will drop out.
Also, note that equation $(3)$ has been shown as an algebraic inequality : so one does not need to verify it by actually checking it for $n = m+1$, $n = m+2$, ... : which is precisely the content of the first part of this answer.
In conclusion, since for all $\epsilon > 0$ we have located a suitable natural number $m$ such that the limit condition holds, we conclude that $L = 0$ is the limit of $u_n$.

For example, let $\epsilon = 0.1$. Then, from above you get that any $m > \frac 1{0.01} = 100$ works. For example, take $m = 101$. For every $n \geq 101$ then $|u_n - L| = |\frac 1{n} - 0| = \frac 1{\sqrt n} < 0.1$ since $n > \frac 1{0.01}$. This algebraic inequality holds for all $n \geq 101$, without having to explicitly check that it works for $n=102,n=54654$ etc.

I hope that this answer at least tells you the following :


*

*How an inequality of variables leads to the direct proof of  possibly infinitely many numerical inequalities, based on the values which the variables can take.

*How to make a decent guess of $L$, by observing $u_n$ or some subsequences of $u_n$.

*How to find $m$ by deconvoluting the expression $|u_n - L|$.
Later on, you will learn better techniques for the latter two points, and then you will learn theorems which will give you limits without needing to proceed by definition. Finally, when a limit does not exist, you will be able to tell by seeing subsequences which go to different numbers, although that will the content of an answer covering an example of a limit not existing.
