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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?

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  • $\begingroup$ You might want to look at the series $\sum_n { n^2 \over 2^n}$. $\endgroup$
    – copper.hat
    Dec 17, 2014 at 18:46
  • $\begingroup$ its cgs by root test. but what we are going to do with the [email protected] $\endgroup$
    – David
    Dec 17, 2014 at 18:47
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    $\begingroup$ Show that the sequence is Cauchy. $\endgroup$ Dec 17, 2014 at 18:49
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    $\begingroup$ $|a_n-a_m| \le |a_n-a_{n-1}| +| a_{n-1}-a_{n-2}| + \cdots +|a_{m+1}-a_m|$. $\endgroup$
    – copper.hat
    Dec 17, 2014 at 18:49
  • $\begingroup$ thank you so much.. it is cauchy so it is converges.. $\endgroup$
    – David
    Dec 17, 2014 at 18:52

2 Answers 2

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\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.

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

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