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Let $a_n $ be sequence of real numbers such that $|a_{n+1}-a_n|\le \frac{n^2}{2^n} $for all n $\in $ N. Then

  1. $ a_n $ is convergent
  2. $ a_n $ is bounded but not convergent
  3. $ a_n $ has 2 limit points.

When it tried this problem, I only thought that given the condition the difference between consecutive terms of sequence is decreasing as we increase n, since $\lim \frac{n^2}{2^n}$=0. So by cauchy convergence criteria sequence should converge.

But recently I came know that $|a_{n+1}-a_n|$ going to zero as n goes to infinity is not a criteria to ensure convergence of sequence. So now how to solve this problem. The answer is option 1.

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  • $\begingroup$ $a_2 = a_1 + (a_2 - a_1)$, $a_3 = a_1 + \dotsc$ $\endgroup$ – Daniel Fischer Jan 24 '15 at 12:57
  • $\begingroup$ For sufficiently large $n$ $$\dfrac{n^2}{2^n}\le\dfrac{1}{(\sqrt{2})^n}.$$ $\endgroup$ – Bumblebee Jan 24 '15 at 13:15
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The answer is only $1$. The inequality implies that $|a_{p} - a_q|$ is smaller that a partial sum of a converging series, namely the series of term $n^2 / 2^n$, so that your sequence is Cauchy, and therefore converges, as $\mathbf{R}$ is complete.

Obviously $a_{n+1} - a_n$ converging to zero is not a criterion, see $ a_n = \ln (n)$.

Details. If $n\geq m$ you have : $a_n- a_m = a_n - a_{n-1} + \ldots + a_{m+1} - a_m$ so that $|a_n- a_m| \leq |a_n - a_{n-1}| + \ldots + |a_{m+1} - a_m| \leq \sum_{k = m}^{n-1} \frac{k^2}{2^k}$. As the series $\sum_kfrac{k^2}{2^k}$ is convergent, for each $\varepsilon > 0$ there exist an $N\in\mathbf{N}$ such that $n\geq m\geq N$ implies that $\left|\sum_{k = m}^{n-1} \frac{k^2}{2^k}\right| \leq \varepsilon$, and therefore, for each $\varepsilon > 0$ there exist an $N\in\mathbf{N}$ such that $n\geq m\geq N$ implies that $|a_n- a_m| \leq \varepsilon$ by the previous majoration. This fact is the definition of $(a_n)_n$ being a Cauchy sequence. But $\mathbf{R}$ is complete, meaning by that that all Cauchy sequences in $\mathbf{R}$ converge, so that $(a_n)_n$ converges, so you have (1). The sequence being convergent, it is bounded, but you don't have (2). Neither have you (3), as being a limit point of a sequence means being limit of a subsequence of the sequence and as if a sequence converges to $l$, so do all its subsequences.

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  • $\begingroup$ Your answer sounds right, but looks foreign to me. Can u decompose it into elementry terms? $\endgroup$ – Foggy Jan 24 '15 at 13:05
  • $\begingroup$ I m not aware of this result that u have stated to show it being cauchy sequence. Can u plz elaborate? $\endgroup$ – Foggy Jan 24 '15 at 13:09
  • $\begingroup$ @User786 Edited my answer with more details $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 24 '15 at 13:13
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You can use the result

If $\sum_n |a_{n+1}-a_n|<\infty$ then $ \left\{a_n \right\} $ is Cauchy.

See here.

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Cauchy is a good way to go, but using the fact that the difference is really small (the difference between consecutive terms does not only go to zero, which would not be enough: it converges very fast). Bound $\lvert a_{n+m}-a_n\rvert$ for $m\geq 1$ and conclude with Cauchy convergence.

If not by Cauchy (although it's, in the end, equivalent), you can use series if you are more comfortable with them. The series $\sum_{n=1}^N(a_{n+1}-a_n)$ is uniformly convergent (why? Recall that $\sum_{n} \frac{n^2}{2^n}$ is a convergent sequence), and that allows you to conclude on the nature of sequence $(a_n)_n$ (why? telescopic series).

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