# 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?

• What is $a$ here?
– S. G
Jan 14 at 15:26
• I forgot to put the restriction of a, however the only thing that is indicated is that $|a|<1$ Jan 14 at 15:36

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, $$\displaystyle n^2 r^n < M^2 r^M < \epsilon.$$