<s>[INST] Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\times2^n}\le\frac{9}{4}-\frac{n^2-2}{2\times2^n}\le\frac{9}{4}$$
Which is true only for $n>1$.
 [/INST] For another proof, you can see that $\displaystyle \frac{u_{n+1}}{u_n}= \frac{1}{2} \left( 1+ \frac{1}{n+1} \right)^2 \leq 1$ with $n\geq 2$ and $\displaystyle u_n= \frac{(n+1)^2}{2^n}$. So $(u_n)$ decreases for $n\geq 2$ whence $\displaystyle u_n \leq u_2=\frac{9}{4}$ for $n \geq 2$. Finally, you only have to check the case $n=1$.
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