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I am having difficulties proving the sequence $x_n = \frac{2^n}{n^2}$ is unbounded.

I can prove it is monotone from some term, but I don't know if it is necessary:

$\frac{2^n}{n^2}$ v $\frac{2^{n+1}}{(n + 1)^2}$

After transformations I get:

$\frac{2^n}{n^2}$ < $\frac{2^{n+1}}{(n + 1)^2}$, when $n > 1 + \sqrt{2}$

However, when I am trying to prove by contradiction that the sequence is unbounded, I get:

$\frac{2^n}{n^2} < M$

$n < log_2 (M*n^2)$

And after that I can only get the initial statement and I do not know how to proceed. Can someone show me what to do next? Thanks!

qex
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2 Answers2

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Notice that $\frac{x_{n+1}}{x_n} = \frac{2}{(\frac{n+1}{n})^2}$ and the ratio approximates 2 for big $n$. Therefore the sequence is not bounded, since for sufficiently big $n$, $x_{n+1} \approx 2 x_n$

Zanzag
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  • Thanks, that was really helpful, but could you please tell me how I prove it more formally. Like, can I say, that since for any number $M$ ($x_n < M$) there exists $k$, such that $2^k * x_n > M$, the sequence is unbounded. Or is it not enough? – qex Sep 06 '21 at 19:36
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    Since $\frac{x_{n+1}}{x_n}$ is an increasing sequence and converges to $2$ we can find and $M$ such that $x_{n+1} \geq \frac{3}{2} x_n$ for every $n > M$. Therefore $x_n \geq \frac{3}{2}^{n-M} x_M$ for every $n > M$ and the sequence is clearly not bounded. – Zanzag Sep 06 '21 at 20:05
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As an alternative we can easily prove by induction that $2^n\ge n^3$ then

$$x_n = \frac{2^n}{n^2}\ge \frac{n^3}{n^2}=n$$

which is clearly unbounded.

Refer to the related:

user
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