3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You might want to look at the series $\sum_n { n^2 \over 2^n}$. $\endgroup$ – copper.hat Dec 17 '14 at 18:46
  • $\begingroup$ its cgs by root test. but what we are going to do with the series?@copper.hat $\endgroup$ – David Dec 17 '14 at 18:47
  • 1
    $\begingroup$ Show that the sequence is Cauchy. $\endgroup$ – André Nicolas Dec 17 '14 at 18:49
  • 1
    $\begingroup$ $|a_n-a_m| \le |a_n-a_{n-1}| +| a_{n-1}-a_{n-2}| + \cdots +|a_{m+1}-a_m|$. $\endgroup$ – copper.hat Dec 17 '14 at 18:49
  • $\begingroup$ thank you so much.. it is cauchy so it is converges.. $\endgroup$ – David Dec 17 '14 at 18:52
3
$\begingroup$

\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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.