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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

mechanodroid
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    Do you know that a convergent serie is Cauchy ? Since $\frac{n^2}{4^n}\leq \frac{2^n}{4^n}=\frac{1}{2^n}$, you should be able to conclude quickely. – Surb Sep 15 '18 at 19:04
  • yes I know that every Cauchy sequence is convergent@Surb –  Sep 15 '18 at 19:24
  • what about the choice of $N$ ..... or I neednot use it?@Surb –  Sep 15 '18 at 19:27
  • So is my above choice of N correct?@Surb –  Sep 15 '18 at 20:14
  • So the limit is 2? this is not so much clear for me because we are beginning from $k = n+1$....... and I think the summation in your last comment is on $k$ not $n$.... am I correct? @Surb –  Sep 16 '18 at 01:27

2 Answers2

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Cauchy means that it converges. Now if we take the fraction criterion of the sequense $a_n=\frac{n^2}{4^n}$ we see that that it is $\lim\frac{a_{n+1}}{a_{n}}=1/4$. It is less than $1$ so the criterion says that it converges i.e it is Cauchy.

dmtri
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  • "Cauchy means that it converges". Of course not ! – Surb Sep 15 '18 at 19:09
  • In a complete metric space like $\Bbb{R}$, Cauchy and convergent are equivalent. – dmtri Sep 15 '18 at 19:12
  • @Surb, Is it true ?or am I mistaken ? Thanks for the feedback. – dmtri Sep 15 '18 at 19:18
  • In a complete space, indeed Cauchy $\iff$ converge (i.e. Cauchy sequence and convergent sequence are equivalent), but you don't define a sequence to be Cauchy as a sequence that converge. So no, they are not the same. – Surb Sep 15 '18 at 20:14
  • @Surb, that is why I wrote "Cauchy means..." and I think it is obvious that we are in the metric space $\Bbb{R}$. – dmtri Sep 16 '18 at 01:49
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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.

mechanodroid
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