I'm working on these supremum and infimum problems and I just want to check to see if my approach is valid. The question asks to find the supremum and infimum of the set $\{\,(n+1)^2/2^n\mid n\in \mathbb{N}\,\}$.
My approach is to take the derivative of the equation and set it equal to $0$. Doing this I find that $n=-1$ and $n=\frac 2 {\log 2}-1$. I also found that $n=\frac 2 {\log 2}-1$ produces a local maximum and $n=-1$ produces a local minimum. Is it accurate to say that these are my infimum and supremum values or not?
I will take any help I can get. Thanks!
Best Answer
Yes, your approach is correct but, since $n \in N$ we need to modify it a little bit.
so $f(x) = \frac{(x+1)^2}{2^x}$ if first increasing and then decreasing(for $x $ non negative), this you check by taking derivative. So for $x \to \infty$, $f(x) $ goes to zero, so inifimum is $0$ .
For supremum your approach is correct, but $x$ shoud be integer, so just find the value of $x$ that is integral. precisely check $x_1 = \left[\frac{2}{\log2}-1\right]$ and $x2 = x1+1$. So $\frac{2}{\log2} = 2.88,\ x_1 = 1, \ x_2 = 2$ so for x2 we get the maximum value of $2.25$.