Prove that $\left(\frac{n+1}{n^2}\right)_{n=1}^{\infty}$
Workings:
Suppose $\epsilon > 0$
Want $N$ = ____ such that
$\left|\frac{n+1}{n^2} - \frac{m+1}{m^2}\right| < \epsilon$ $(\forall n,m > N)$
$\left|\frac{n+1}{n^2} - \frac{m+1}{^2m}\right|$
$\leq\left|\frac{n+1}{n^2}\right| + \left|\frac{m+1}{m^2}\right|$
$= \frac{n+1}{n^2} + \frac{m+1}{m^2}$
$= \left(\frac{1}{n} + \frac{1}{n^2}\right) + \left(\frac{1}{m}+\frac{1}{m^2}\right)$
Now I am not too sure on hwat to do. Any help will be appreciated.
To continue, write
$$\frac{1}{n} + \frac{1}{n^2} + \frac{1}{m} + \frac{1}{m^2} \le \frac{1}{n} + \frac{1}{n} + \frac{1}{m} + \frac{1}{m} = \frac{2}{n} + \frac{2}{m}.$$
Then note that given $\epsilon$, setting $N > \frac{4}{\epsilon}$ will make $\frac{2}{n} + \frac{2}{m} < \epsilon$ for all $n \ge N$. Thus $$\left|\frac{n + 1}{n^2} - \frac{m + 1}{m^2}\right| < \epsilon$$ for all $n \ge N$.