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


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?

  • 1
    $\begingroup$ What is $a$ here? $\endgroup$
    – S. G
    Jan 14 at 15:26
  • $\begingroup$ I forgot to put the restriction of a, however the only thing that is indicated is that $|a|<1$ $\endgroup$ 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 $~\displaystyle \left[\frac{C+1}{C}\right]^2 \times r < 1.$

$r^C < \epsilon \implies C^2 r^c < \epsilon ~C^2.$

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


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