<s>[INST] Is the sequence $a_{n} = 1 + \frac14 + \frac{2^2}{4^2} + \cdots +\frac{n^2}{4^n}$ Cauchy? I think that it is Cauchy (but I am not sure of this) and this is my proof:
$$|a_{m} - a_{n}| = \left|\frac{n+1}{4^{n+1}} + \frac{n+2}{4^{n+2}} + ..... + \frac{m^2}{m}\right| =\sum_{k=n+1}^{m} \frac{k^2}{4^k}$$ 
And then knowing that $4^n \geq n^2$ for all $n \geq 4$. (which I will prove by induction)
Then $a_{m} - a_{n} =\sum_{k=n+1}^{m} \frac{k^2}{4^k} \le \sum_{k=n+1}^{m} \frac{k^2}{k^2} \text{(for $n \geq 4$)} = m-n  < m$ and we want $a_{m} - a_{n} < \epsilon$, then we want $m<\epsilon$, but $m<\epsilon$ implies $n<\epsilon$ because $m\geq n$. So can I choose $N =  \lfloor\epsilon\rfloor +1 $?
Is this a correct choice that shows that my sequence is Cauchy?
I hope that my question is following the instructions of your esteemed, educational and useful site  
 [/INST] Prove by induction that $k^2 \le 2^k$ for $k \ge 4$ so 
$$|a_{m} - a_{n}| \le \sum_{k=n+1}^m \frac{k^2}{4^k} \le \sum_{k=n+1}^\infty \frac1{2^k} = \frac1{2^{n}}\xrightarrow{m,n\to \infty} 0$$
Hence $(a_n)_n$ is Cauchy.
</s>