Q:

Convergence of sequence s.t $|a_{n+1}-a_n|\le \frac{n^2}{2^n}$

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

$ a_n $ is convergent
$ a_n $ is bounded but not convergent
$ 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.

A:

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.

