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I'm required to prove $a_{n}=\frac{n+1}{\sqrt{n}}$ diverges by using the negation of the definition of convergence, that is:

$(\forall\epsilon >0)(\exists m \in \mathbb{N})$ such that $\forall (n>m) \in \mathbb{N}\:\:\left | f_{n}-L\right |<\epsilon$

The negation of this is: $(\exists \epsilon>0)(\forall m \in \mathbb{N})$ such that $\exists(n>m)\in \mathbb{N}\:\: \left | f_{n}-L\right |\geq \epsilon$

Now I've done proofs like this before for oscillatory sequences like $a_{n}=(-1)^{n}$ where you can split it up into two cases, one for $L<0$ and one for $L \geq 0$ and then choose n to be an arbitrary even or odd number greater than m respectively and then work straight from the definition, so this is what I tried first.

For $L\leq0$ I chose $n_{0}=4m^{2}-1>m$ and $\epsilon = 1$

Then $\left | f_{n_{0}}-L\right |=f_{n_{0}}+\left | L\right |=2m+\left | L\right |\geq2>\epsilon=1$

So for every natural number there exists a natural number n such that the the distance between the n'th term of $f_{n}$ and any $L\in\mathbb{R}$ where $L<0$ so $f_{n}$ doesn't converge to any negative limit.

I'm not 100% confident with this, but I think its mostly okay, its the $L>0$ case that I'm completely lost for, I'd really appreciate some help if possible. Thanks!

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Hint: Divide the sequence into two parts: $$a_n=\frac{n}{\sqrt n}+\frac{1}{\sqrt{n}}=\sqrt n+\frac{1}{\sqrt{n}}$$ Now you get a term which is convergent, and another one which should be easy to check that diverges. Let's foccuss on the first term, which is divergent $g_n=\sqrt{n}$

Edited

Let's imagine that $\exists L,\epsilon\in \mathbb{R},m\in \mathbb{N} $ such that $|g_m-L|<\epsilon$. For considering $L$ the limit of the sequence, the sequence has to satisfy $\forall(n>m)\in\mathbb{N}\;|g_n-L|<\epsilon$. For demonstrating that it is not of the sequence is enough finding one $n>m$ such $|g_n-L|\ge\epsilon$. Let's take for example $n=2m$ and $g_{2m}=\sqrt{2}g_m$. Now, going back to the inequality, the condition reads $$|\sqrt2 g_{m}-L|<\epsilon\to\left|g_m-\frac{L}{\sqrt{2}}\right|<\frac{\epsilon}{\sqrt{2}}$$ you can call $L/\sqrt2=L'$ and $\epsilon/\sqrt2=\epsilon'$. This condition would mean that $L'$ is also a limit of the sequence, what it is impossible unless $L=L'=0$. And you can demonstrate that $L=0$ is also not a limit, using a similar reasoning.

Strictly speaking, the reasoning before shows that the sequence has no real limit. But it can be demonstrated that the limit is actually infinity. Similarly to your definition of limit, We know that the limit is infinity if $\forall \epsilon\in\mathbb{R}^+$ such that $\forall n>m,\;|g_n|>\epsilon$. This definition will imply that the sequence is unbounded. Firstly, you can determine $m$ $$|g_m|=\sqrt{m}>\epsilon\to m=\epsilon^2$$ now, if you take any $g_n$ with $n>m$ you know that $g_n=\sqrt{n}>g_m>\epsilon$. Then, the limit is infinity

Finally, the full sequence is a sum of two: one convergent and one divergent, so the full one will be divergent.

Please, let me know if I missed something or the answer is not clear enough!

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  • $\begingroup$ I'm really sorry, I've been working at this for a solid two hours more and I still can't see how to proceed. I just haven't seen any examples of proofs of this kind for sequences which diverge to infinity so I'm finding it really hard to construct one. I think I could manage it by contradiction or maybe induction, but I just cant seem to figure out how to do it strictly by the above definition. Could you possibly elaborate or maybe show me how to get started? I tried splitting it up, but again couldn't seem to make it conform to the definition. thanks heaps for your help! $\endgroup$ – CoffeeCrow Apr 21 '16 at 12:33
  • $\begingroup$ @CoffeeCrow I have included some more details. Please, let me know if the proof is uncomplete, or if there is something missing $\endgroup$ – seoanes Apr 21 '16 at 13:04
  • $\begingroup$ So if I understand correctly this is pretty much a proof by contradiction, showing that if the sequence isn't divergent it would have to converge to two different limits? This makes sense and I think I can follow the reasoning, but I still don't see how to prove this using just the negated definition of convergence. $\endgroup$ – CoffeeCrow Apr 21 '16 at 13:53
  • $\begingroup$ Check out my last edit. I think it is more in the philosophy of what you are looking for, but I am not sure that it is completely the same you are asking $\endgroup$ – seoanes Apr 21 '16 at 14:07

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