Is the sequence ($x_n$)= $a^n(n^2)$ , $0<a<1$ convergent?
Proof that the sequence is unbounded and hence divergent:
Assume it is bounded, then there exists $M>0$, such that $M\geq|x_n|$ for all $n \in \Bbb N$. We note that since $a>0$ and $n>0$, all the terms of the sequence are positive. So $M\geq x_n$. Since $a^n$ is positive, then its reciprocal must also be positive. Hence by multiplying both sides by $1/a^n$, we obtain $M/a^n\geq n^2$ for all $n \in \Bbb N$.
By the Archimedean property, there exists $n_1 \geq M/a^n$, since $M/a^n$ is positive. Let $n_2 = n_1 +1$. Then $n_2$ is a natural number, and $n_2 > M/a^n$, so $$ n_2^2 > n_2 M/a^n > n_1 M/ a^n \geq M/ a^n $$ so $n_2^2 > M/ a^n$. But then $M/a^n \geq n_2^2$ since $n_2 \in \Bbb N$. Since $n_2^2$ cannot be more than $M/a^n$ and less than or equal to $M/a^n$ at the same time, we have derived a contradiction and hence the sequence is unbounded.
Proof that the sequence converges: We employ the Cauchy ratio test.
Take the ratio $x_{n+1} / x_n$. We obtain $$ (a^{n+1} (n+1)^2)/a^n n^2 = a(n+1)^2 / n^2 = a(n^2/n^2 +2n/n^2 + 1/n^2) = a(1+2/n+ 1/n^2).$$ Since the sequences $1/n^2$ and $1/n$ both converge to $0$, the sequence converges to $a(1)+ a(2)(0)+a(0)= a$. Since $a<1$, then the sequence converges.
Could someone help me, which of these proofs is false? Am I making any mistakes, any areas I can improve on? Thank you!
