We can prove by induction that $a_n=\frac{\binom{2n}{n}}{4^n}$ for all $n\in \mathbb{Z}^+$. For the second part, we can prove, again, by induction that $a_n\leq \frac{1}{\sqrt{3n+1}}$.

ThE-dArK-lOrD wrote: We can prove by induction that $a_n=\frac{\binom{2n}{n}}{4^n}$ for all $n\in \mathbb{Z}^+$. For the second part, we can prove, again, by induction that $a_n\leq \frac{1}{\sqrt{3n+1}}$. I don’t understand your second part’s solution? Can you explain clearly,plz?

We can figure out that a_1=1/2, a_2=3/8, a_3=5/16, a_4=35/128.It's clear that 2na_n=(2n-1)a_(n-1), and it's quite easy to prove it by induction.

bumppppp