Given is a sequence of two-dimensional vecotrs $bn= (\frac{1}{n}, \frac{1}{n^2})^T$
For any $\epsilon \gt 0$, find a condition for N such that $||bn||\le \epsilon, ∀n \ge N$ is guaranteed.
Hint: $ {\sqrt {a+b}}\le {\sqrt a+\sqrt b}$ and $ |a|= {\sqrt {a^2}} $.
I am not sure if I am right, but I have calculated $||bn||$ using $ ||v||= \sqrt {v_1^2+v_2^2+...+v_n^2} $.
So:
$||bn||\le \epsilon$
i.e.
$ \sqrt {\frac{n^2+1}{n^4}}\le \epsilon$
After a few steps I have obtained:
$ {n^2}\ge {\frac{1+\sqrt {1+4\epsilon^2}}{2\epsilon^2}}$
There are two possible solutions, but the negative one cannot be accepted, so:
$ n\ge \sqrt \frac{1+\sqrt {1+4\epsilon^2}}{2\epsilon^2} $
I don't know where I am supposed to use the hint. Shoud I write $ n\ge \sqrt \frac{1}{2\epsilon^2}+\sqrt \frac{\sqrt {1+4\epsilon^2}}{2\epsilon^2} $?
Any help is appreciated.
Thanks in advance.
Note that$$\lVert b_n\rVert=\frac{\sqrt{n^2+1}}{n^2}\leqslant\frac{n+1}{n^2}=\frac1n+\frac1{n^2}\leqslant\frac2n.$$So, take $N\in\mathbb N$ such that $\frac2N\leqslant\varepsilon(\iff N\geqslant\frac2\varepsilon)$ and then $n\geqslant N\implies\lVert b_n\rVert<\varepsilon$.