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I am considering how to find a supremum and infimum of $\frac{(n+1)^{2}}{2^{n}}, n\in \mathbb{N}$. I proved with induction that for $n\geq 6$ it goes to zero so infimum is also zero. For other n $(=1, 2, 3, 4, 5)$ I checked that supremum is $\frac{9}{4}$. And now I would like to prove it with Archimedes rule using $\varepsilon $, but I end with $\frac{9}{4}- \varepsilon < \frac{(n+1)^{2}}{2^{n}}$, and I have no idea how to transform it to use Archimedes; something like $n> ...$ Any idea? Thanks in advance!

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2 Answers 2

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To find the sup of your sequence, use the first derivative test which will give you a hint for finding it. In this case, consider the function

$$ f(x) = \frac{ (x+1)^2 }{2^x}\,. $$

Taking the first derivative and equating it to $0$ to find the stationary points. Doing that, you will get the two points $x=-1,x=\frac{2-\ln(2)}{\ln(2)}=1.885390081$. Since, we are interested in $x\geq 0, $ we will pickup $ x=1.885390081. $ Checking the second derivative at this point, we have $ f'' < 0. $ which means that $x=1.885390081$ gives the max of the function. So this gives the hint where the sup of your sequence is.

For the inf of the sequence, just prove that the sequence decreases to $0$ for $n>2.$

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Let $u_n=(n+1)^2/2^n$, then $u_{n}/u_{n-1}=(n+1)^2/(2n^2)$ hence $u_{n}\gt u_{n-1}$ if and only if $(n+1)^2\gt2n^2$ if and only if $n^2-2n-1\lt0$ if and only if $(n-1)^2\lt2$ if and only if $-0.4^-=1-\sqrt2\lt n\lt1+\sqrt2=2.4^+$. Thus $u_0\lt u_1\lt u_2$ and $u_2\gt u_3\gt u_4\gt\ldots$ and $u_n\gt u_{n+1}$ for every $n\geqslant2$.

The supremum of $(u_n)_{n\geqslant0}$ is $u_2=9/4$ (which is also the maximum). The infimum is either $u_0$ or the limit of $(u_n)_{n\geqslant0}$ at infinity. Since $u_0=1$ and $\lim\limits_{n\to\infty}u_n=0$, the infimum is zero (and there is no minimum).

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