how to solve linear and exponential mixed equation $\frac{(n+1)^{2}}{2^{n}}$

Question

this equation is actually a part of a bigger question. if says determine $sup (A)$ and $inf (A)$ where A is the set of $\frac{(n+1)^{2}}{2^{n}}$

now my problem is not $sup (A)$ and $inf (A)$

since at some value $n $ (which I found by repetitive calculation to be $n \gt 6$ thus )the elements of the set will tend to $0$ after $n = 6$ which is fine for the inf (A)

my question is how to determine this n without trail and error

and

if $n$ was big, how to determine the $sup (A) $ (pre-assuming that for $n \lt 6$, or other value, the value of $x$ in $A $ is greater than $1$

Answer 1

You can set $f(x)=\dfrac{(x+1)^2}{2^x}$.

$f'(x)=\underbrace{-\dfrac{(x+1)}{2^x}}_{<0}\times\big(\ln(2)(x+1)-2\big)$

Since we are in fact interested in natural numbers, then $x\ge 0$, allowing to find the sign of the derivative easily.

There is an extremum point at $x_0=\frac{2}{\ln(2)}-1\approx 1.88$ and $f'>0$ to the left, $f'<0$ to the right, so $x_0$ is a global maximum on $[0,+\infty)$

However since $x_0$ is not an integer, the maximum at integer points is either $f(1)=2$ or $f(2)=\frac 94$ (for the two integer abscissa surrounding $x_0$).

Thus $\sup(A)=\frac 94$ and is reached for $n=2$.

Answer 2

hint

For the inf,

For all $n $

$$0 < \frac{(n+1)^2}{2^n}$$ and

$$\lim_{n\to +\infty}(n+1)^2e^{-n\ln(2)}=0$$ since the exponential is faster than the power.

thus $\inf A=0$.

For the sup, you should study the function $x\mapsto \frac{(x+1)^2}{2^x}$.