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I think I have a proper understanding of the general procedure, but I'm having difficulty manipulating my inequality so that I can isolate $n$ by itself. Sadly I wasn't given many examples to model my answer on.

Prove that $\displaystyle\lim_{n\to\infty}\frac{n+1}{n^2+1}=0$

So I'm given $L=0$. I then look at the inequality

$$\left| \frac{n+1}{n^2+1}-0\right|<\epsilon$$

but I have no idea how to isolate $n$. The best I can come up with, which may be the right idea, is to use another function $f$ such that


and then work with that. But my idea of using $f=\lvert n+1\rvert$ seems to have me a bit stuck too.

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Nowhere in your start of the proof do you use $N$. Do you understand that $N$ and $n$ are different in $\epsilon-N$ proofs? – Thomas Andrews Sep 28 '13 at 15:21
Given $\epsilon>0$ you need an $N$ such that for all $n>N$, $$\left|\frac{n+1}{n^2+1}\right|<\epsilon$$ – Thomas Andrews Sep 28 '13 at 15:26
Also, when writing a function $f$, you really ought to write :$$\frac{n+1}{n^2+1}<f(n)<\epsilon$$ Writing $f<\epsilon$ makes $f$ look like $f$ is a constant... – Thomas Andrews Sep 28 '13 at 15:36
I was aware of those things, actually.. I was just trying to be concise in my question. I was getting specifically to the point I was having trouble with... but I think I solved it thanks to one example I found buried deep in my textbook. I posted it below. – agent154 Sep 28 '13 at 16:18
Well, one reason to write longer is to make clear what you understand. People on M.SE have a lot of different experience levels, so when you skip crucial stuff or use odd notation, we have to guess if you understand it. Verbosity is not a crime. Indeed, it leads to better answers because people can tell what you know and what you don't know. Nothing is aided by writing $f$ instead of $f(n)$, or by confusing $n$ and $N$ in your text. – Thomas Andrews Sep 28 '13 at 16:29

Since the given sequence is positive for all $n\geq 1$, we can drop the absolute value signs.

Consider the inequality: \begin{equation*} \frac{n+1}{n^2+1}<\epsilon \end{equation*} This is a messy inequality, as solving for $n$ would be rather difficult. Instead we shall find an upper bound for the numerator $(n+1)$ and a lower bound on the denominator $(n^2+1)$ so that we can construct a new sequence $f_{n}$ such that $$\frac{n+1}{n^{2}+1}<f_{n}<\epsilon.$$

Since $n+1$ behaves like $n$ at very large numbers, we shall try to find a value for $b$ such that $bn>n+1$. And since $n^{2}+1$ behaves like $n^{2}$ for very large numbers, we shall try to find a value for $c$ such that $cn^{2}<n^{2}+1$. $n+1<n+n=2n$ and $n^{2}+1>\frac{1}{2}n^{2}$ for all $n\geq 1$.

Now, consider the new inequality: \begin{align*} \frac{n+1}{n^{2}+1}<\frac{2n}{\frac{1}{2}n^{2}}<\frac{4}{n}<\epsilon \end{align*} We have \begin{equation*} \frac{4}{n}<\epsilon\iff \frac{4}{\epsilon}<n \end{equation*}

Let $\displaystyle N(\epsilon)=\lfloor4/\epsilon\rfloor$. Therefore, $\forall \epsilon>0,\ \exists N(\epsilon)=\lfloor4/\epsilon\rfloor\ni\forall n>N(\epsilon)$ \begin{align*} \frac{n+1}{n^{2}+1}&<\frac{4}{n}<\frac{4}{N(\epsilon)+1}<\frac{4}{4/\epsilon}=\epsilon \end{align*}

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Yes, this works. – Thomas Andrews Sep 28 '13 at 16:24
+1 good solution. Using the lower bound $n^2 + 1 > n^2$ in stead of $n^2 + 1 > \frac12 n^2$ would be more efficient, but it doesn't really matter. – 6005 Sep 28 '13 at 16:24
Thanks. I was actually concerned about that myself, because it seemed my prof was writing that on the board... or maybe I thought he did, as I was writing it in my notes. – agent154 Sep 28 '13 at 16:24

Hint: Show that $0<\frac{n+1}{n^2+1}<\frac{1}{n-1}$.

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I'm sorry - I really don't understand what I'm supposed to do based on that hint. This is my first time trying to do this on my own so I don't have much of a foothold on the procedure yet. – agent154 Sep 28 '13 at 15:09
Sorry, I always try to start with a hint. – Thomas Andrews Sep 28 '13 at 15:34
Maybe you should try to prove that $$\lim_{n\to\infty}\frac{1}{n-1} =0?$$ – Thomas Andrews Sep 28 '13 at 15:45
Does my proposed solution below work? – agent154 Sep 28 '13 at 16:21
Yes, your solution works. In my solution $N(\epsilon)=1+\lfloor 1/\epsilon\rfloor$. So if $n>N$ then $n-1>1/\epsilon$ so $\frac{1}{n-1}<\epsilon$. – Thomas Andrews Sep 28 '13 at 16:26

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