Finding an $N$, such that $\forall n \geq N$... Ok, so I am stuck regarding the following question;
Let $\epsilon > 0$ be a positive real number. Find a natural number $N \in$ $\mathbb{N}$ such that $$ |a_n| =\left|\frac{n + \frac{1}{n}}{n+1} - 1\right| < \epsilon, \forall n \geq \mathbb{N}$$
Now, my thinking when I'm posed with a question like this is to try and rearrange to get $n > f(\epsilon)$, then one can easily deduce what the value of $N$ should be, for instance by taking the integer part of $f(\epsilon)$ and adding $1$, which ensures that it is an integer and also ensures that it is greater or equal to the original $f(\epsilon)$.
Now, with an expression like this, I cannot see how to possibly rearrange it to get the required expression, but I have been told that if I find an $f(n) \geq |a_n|$, then I can use the $N$ I get from this as the $N$ for the original expression. This intuitively makes sense as this $f(n)$ is larger, and so you'd need a larger $N$ for it to be smaller than $\epsilon$, and so this 'larger $N$' would also work for the original expression.
But I run in to trouble with finding this 'nice $f(n)$', I mean can I take any larger $f(n)$? For example $|a_n| \leq \left|\frac{n(n+1)}{n+1} - 1\right| = |n - 1|$. Because $n^2 \geq n \enspace,    \forall n \geq 1$ and $n \geq \frac{1}{n} \enspace\forall n \geq 1$.
But this then gives $n < \epsilon + 1$ which is nonsense so I'm not sure if my method is correct or not.

HINT: Note that
$$\left|\frac{n+\frac1n}{n+1}-1\right|=\left|\frac{\frac1n-1}{n+1}\right|=\frac{1-\frac1n}{n+1}<\frac1{n+1}\;.$$
If you can get $\frac1{n+1}<\epsilon$, then you automatically get $|a_n|<\epsilon$.
