# 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?

• You might want to look at the series $\sum_n { n^2 \over 2^n}$. – copper.hat Dec 17 '14 at 18:46
• its cgs by root test. but what we are going to do with the series?@copper.hat – David Dec 17 '14 at 18:47
• Show that the sequence is Cauchy. – André Nicolas Dec 17 '14 at 18:49
• $|a_n-a_m| \le |a_n-a_{n-1}| +| a_{n-1}-a_{n-2}| + \cdots +|a_{m+1}-a_m|$. – copper.hat Dec 17 '14 at 18:49
• thank you so much.. it is cauchy so it is converges.. – David Dec 17 '14 at 18:52

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 $n,m\ge N'$, $$|a_{n}-a_m| \le |a_{n}-a_{n+1}| + |a_{n+1}-a_{m+1}|+|a_{m+1}-a_m|< 3\frac{\epsilon}{3}=\epsilon$$