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$a_n=\frac{n+1}{n+2}$.

I always get stuck towards the endpoint, as I don't really know what to do.

Here goes: let $\epsilon>0$ be given. In order for $a_n$ to converge to 1 (which I've informally come up with for the limit), we must show that $\forall N \geq n$, $\left | a_n-1\right |<\epsilon$.

Now, $\left | a_n-1\right |=\left | \frac{n+1}{n+2}-1\right |=\left | \frac{1}{n+2} \right |=\frac{1}{n+2}$.

But then where do I go from here? Can I just say that in order for $\frac{1}{n+2}<\epsilon$, we must have that $n+2>\frac{1}{\epsilon}$, so we need that $n>\frac{1}{\epsilon}-2$.

So, let $N=\frac{1}{\epsilon}-2$, then $a_n$ converges. Is that the end of the proof? If this is completely wrong, could you please give me a general 'recipe' on how to approach these 'prove from first principles' problems, so that I can apply this technique to other sequences? Algebraically, I'm fine with everything until the $=\frac{1}{n+2}$ part. After that I'd be lying if I didn't say that I don't have a clue what's going on. As much detail in the technique as possible (in the general case for any sequence) would be greatly appreciated.

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If you let $\epsilon = \frac{1}{n}$, then you'll have $\frac{1}{n+2} < \frac{1}{n}$ for all $n$. Then you're done, as it is implies that $\left| a_n -1 \right| < \epsilon$.

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  • $\begingroup$ Apart from that, is the rest ok? And what about the general case? $\endgroup$ – beep-boop Feb 14 '14 at 22:21
  • $\begingroup$ So, in all cases, do I just let $\epsilon$ be some appropriate function of $n$? Is my method ok? $\endgroup$ – beep-boop Feb 14 '14 at 22:23
  • $\begingroup$ In my opinion, it is fine. $\endgroup$ – Chris K Feb 14 '14 at 22:25
  • $\begingroup$ You get confused when you let $N = \frac{1}{e} - 2$, which doesn't make sense when you remark that $N$ is a natural number. You're doing fine until you say that "in order for $\frac{1}{n+2} < \epsilon$, we must have..." which is when it all goes downhill. You've picked the $N=1$ it converges to, all you have to do is pick an $\epsilon$ so that $\frac{1}{n+2} < \epsilon$ for all $n \geq N$. $\endgroup$ – Newb Feb 14 '14 at 22:26
  • $\begingroup$ And the way to do that is to define $\epsilon$ in terms of $n$. In our case, the easiest way to do this is just to let $\epsilon = \frac1n$. $\endgroup$ – Newb Feb 14 '14 at 22:27

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