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Verification of proof of nonconvergence

Proof Verification: Cauchy Sequences are convergent.I don't understand the proof that $Bbb{R}$ is complete with respect to Cauchy sequences.Proof verification regarding whether a certain property of a sequence implies that it is Cauchy.Proof verification: convergent series with a finite number of negative terms is Absolutely ConvergentProof Verification: Regarding equivalent sequences and Cauchy sequences, Tao Analysis I: Exercise 5.2.1.A convergent sequence is bounded: Proof VerificationProof verification - Cauchy completeness of $mathbb{R}$Cauchy sequence exercise question (Exercise 2.6.4 Abbott analysis)Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is Cauchy sequenceLet there be a sequence such that the distance between two consecutive terms converges to 0. Must this sequence converge?

0

$begingroup$

I'm not sure if my proof is sound or not, I was wondering if anyone could verify.

Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.

Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$

Any help would be appreciated, thanks!

real-analysis proof-verification convergence cauchy-sequences

share|cite|improve this question

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

$endgroup$

add a comment | 

0

$begingroup$

I'm not sure if my proof is sound or not, I was wondering if anyone could verify.

Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.

Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$

Any help would be appreciated, thanks!

real-analysis proof-verification convergence cauchy-sequences

share|cite|improve this question

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

$endgroup$

add a comment | 

$begingroup$

I'm not sure if my proof is sound or not, I was wondering if anyone could verify.

Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.

Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$

Any help would be appreciated, thanks!

real-analysis proof-verification convergence cauchy-sequences

share|cite|improve this question

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

$endgroup$

share|cite|improve this question

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

share|cite|improve this question

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

edited 6 hours ago

José Carlos Santos

165k22132235

edited 6 hours ago

José Carlos Santos

165k22132235

asked 6 hours ago

B RetnikB Retnik

534

asked 6 hours ago

B RetnikB Retnik

534

2 Answers
2

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1

$begingroup$

That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).

You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.

share|cite|improve this answer

edited 4 hours ago

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

$endgroup$

add a comment | 

1

$begingroup$

You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
$$frac{3n}{n+1}a_n rightarrow 3l$$

But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.

share|cite|improve this answer

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

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add a comment | 

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1

$begingroup$

That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).

You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.

share|cite|improve this answer

edited 4 hours ago

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

$endgroup$

add a comment | 

1

$begingroup$

That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).

You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.

share|cite|improve this answer

edited 4 hours ago

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

$endgroup$

add a comment | 

$begingroup$

That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).

You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.

share|cite|improve this answer

edited 4 hours ago

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

answered 6 hours ago

José Carlos SantosJosé Carlos Santos

165k22132235

1

$begingroup$

You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
$$frac{3n}{n+1}a_n rightarrow 3l$$

But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.

share|cite|improve this answer

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

$endgroup$

add a comment | 

1

$begingroup$

You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
$$frac{3n}{n+1}a_n rightarrow 3l$$

But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.

share|cite|improve this answer

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

$endgroup$

add a comment | 

$begingroup$

You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
$$frac{3n}{n+1}a_n rightarrow 3l$$

But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.

share|cite|improve this answer

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

$endgroup$

share|cite|improve this answer

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112

answered 6 hours ago

TheSilverDoeTheSilverDoe

2,973112