let $(a_n)$ be a sequence of real numbers such that $|a_{n+1}-a_n|\leq \frac {n^2}{2^n}$ for all $n\in \mathbb N$. Then let $(a_n)$ be a sequence of real numbers such that $|a_{n+1}-a_n|\leq \frac {n^2}{2^n}$ for all $n\in \mathbb N$. Then


(a). The sequence $(a_n)$ may be unbounded.
(b). The sequence $(a_n)$ is bounded but may not converge.
(c). The sequence $(a_n)$ has exactly two limit points.
(d). The sequence $(a_n)$ is convergent.


My work
$$|a_n|-|a_{n+1}|\leq |a_n-a_{n+1}|\leq \frac {n^2}{2^n}$$
$$|a_n|\leq |a_{n+1}|+\frac {n^2}{2^n}$$
So, we can conclude that the sequence is either increasing or decreasing. So, sequece may be unbouded. Is it correct?
 A: \begin{align}
|a_{n}-a_m|&\leq |a_n-a_{n+1}|+|a_{n+1}-a_{n+2}|+\ldots+|a_{m-1}-a_m|\\
&\leq \frac{n^2}{2^n}+\frac{(n+1)^2}{2^{n+1}}+\ldots+\frac{m^2}{2^m}\\
&=\frac{n^2}{2^n}\left\{1+\frac{(1+\frac 1 n)^2}{2^{n}}+\ldots+\frac{(1+\frac {1}{m-n})^2}{2^{m-n}}\right\}\\
&\to 0 \text{ as } n\to \infty
\end{align}
Therefore, the sequence is cauchy, so that its converges.
A: Using copper.hat's hint, for all $n,m \in \mathbb{N}$
$$
|a_n-a_m| \le \sum_{i=m}^{n-1}\frac{i^2}{2^i}
$$
Let $\epsilon>0$ be given.
You have noticed that $$\sum_{i=1}^{\infty} \frac{i^2}{2^i}$$
converges. 
This means that the sequence of partial sums:
$$
A_N=\sum_{i=1}^{i=N}\frac{i^2}{2^i}
$$
is Cauchy.
Thus there exists $N'\in \mathbb{N}$ such that for all $m,n \ge N'$:
$$
\left|\sum_{i=1}^{i=n}\frac{i^2}{2^i}-\sum_{i=1}^{i=m}\frac{i^2}{2^i}\right|=\left| \sum_{i=m+1}^{i=n}\frac{i^2}{2^i}\right|<\frac{\epsilon}{3}
$$
Thus: for $n,m\ge N'$,
$$
|a_{n}-a_m| \le |a_{n}-a_{n+1}| + |a_{n+1}-a_{m+1}|+|a_{m+1}-a_m|< 3\frac{\epsilon}{3}=\epsilon
$$
