Given $\epsilon>0$, calculate $m_\epsilon \in \mathbb{N}$ such that for all $n \ge m_\epsilon$ I plan to solve the following exercise
given $\epsilon>0$, calculate $m_\epsilon \in \mathbb{N}$ such that for all $n \ge m_\epsilon$ it is verified that $|x_n-x|<\epsilon$.
In this particular case we have that
$x_n= n^{2}a^{n}$ and $x=0$, also $|a|<1$
It is clear in a way that
$$0 \le x_n=|x_n-0|=|n^{2}a^{n}|< \epsilon$$
Therefore, we can consider that
$$|a^{n}|<\frac{\epsilon}{n^{2}}$$
I am not sure if the above is entirely true, and I have not been able to find the value of m_e requested. Any help on this?
 A: Let $~r = |a| \implies 0 < r < 1.$
Choose $B \in \Bbb{Z^+}$ so that
$\displaystyle r^B < \epsilon.$
Clearly, $~\forall n > B, ~r^n < \epsilon.$
As $~\displaystyle ~n \to \infty, \frac{n+1}{n} \to 1.$
Therefore, as $\displaystyle ~n \to \infty, \left[\frac{n+1}{n}\right]^2 \to 1.$
If $~\displaystyle \left[\frac{B+1}{B}\right]^2 \times r < 1,~$ set $~C = B$.
Otherwise, choose $~C\in \Bbb{Z^+}, C > B,~$
such that $~\displaystyle \left[\frac{C+1}{C}\right]^2 \times r < 1.$
Therefore 
$r^C < \epsilon \implies C^2 r^c < \epsilon ~C^2.$
Set $~\displaystyle d = \left[\frac{C+1}{C}\right]^2 \times r 
\implies 0 < d < 1.$
Choose $~E \in \Bbb{Z^+},~$ such that
$~\displaystyle d^E < \frac{1}{C^2}.$
Set $M = E + C$.
Then $\displaystyle M^2r^M = C^2 r^c \times \prod_{i=1}^E 
\left[\left(\frac{C+i}{C+i-1}\right)^2 \times r\right].$
Therefore,
$~\displaystyle M^2 r^M \leq C^2 r^C \times 
\left[\left(\frac{C+1}{C}\right)^2 \times r\right]^E = C^2 r^C \times d^E.$
Therefore, $~\displaystyle M^2 r^M < \epsilon ~C^2 \times \frac{1}{C^2} = \epsilon.$
For any $~n \in \Bbb{Z^+},~$ such that
$~n > M$:
$\displaystyle n^2 r^n = M^2 r^M \times 
\prod_{i=1}^{n - M} \left[\left(\frac{M + i}{M + i - 1}\right)^2 \times r\right].$
Therefore, $\displaystyle n^2 r^n
\leq M^2 r^M \times \left[\left(\frac{M+1}{M}\right)^2 \times r\right]^{n - M}.$
Therefore, $\displaystyle n^2 r^n < M^2 r^M < \epsilon.$
