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Convergence of sequence using epsilon [closed]

Difference between convergent sequence and convergent subsequenceCauchy sequence convergenceSequence and subsequence convergenceProving a this sequence converges using epsilon definitionThe connection between the supremums of a function sequence, and the supremum of the limit fumctionDisprove a limit using epsilon definitionSteps for proving that a sequence converges, using the epsilon definition of convergenceQuestion regarding subsequences and a bounded sequenceConditions on convergence of a sequenceConvergence from sequence inequality

0

2

$begingroup$

Prove that, for any $ninmathbbN$,$$fracn^22n^3-1$$using an $ε$-$N$ proof.

I am suppose to prove the sequence above converge using $ε$-$N$ proof. However there is no example in my textbook, I am suppose to use what I know about $ε$-$N$ proofs, choosing an appropriate $ε$.

sequences-and-series

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edited Mar 22 at 2:46

Saad

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asked Mar 22 at 0:33

mushimastermushimaster

16510

$endgroup$

closed as off-topic by RRL, Saad, Leucippus, Eevee Trainer, Wouter Mar 22 at 9:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Leucippus, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How come you don't have any examples?
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 0:35
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I'm guessing that you should have a limit somewhere in you question, right?
  $endgroup$
  – D.B.
  Mar 22 at 0:35
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You need to finish your sentence!
  $endgroup$
  – Ted Shifrin
  Mar 22 at 0:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show that $frac1n - 1 to 0$ using $varepsilon$s and $N$s?
  $endgroup$
  – Theo Bendit
  Mar 22 at 0:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Saad, it is always better to type the Greek name of Greek letters (such as epsilon) rather than paste them as characters, because the latter version belongs to font families that maybe not all browsers can display (depending on what fonts a user has installed). Second, people talk about you, and I myself believe that yout behaviour regarding edits is wrong.
  $endgroup$
  – Alex M.
  Mar 23 at 21:38
  

  
  
  
  

 | 
show 3 more comments

0

2

$begingroup$

Prove that, for any $ninmathbbN$,$$fracn^22n^3-1$$using an $ε$-$N$ proof.

I am suppose to prove the sequence above converge using $ε$-$N$ proof. However there is no example in my textbook, I am suppose to use what I know about $ε$-$N$ proofs, choosing an appropriate $ε$.

sequences-and-series

share|cite|improve this question

edited Mar 22 at 2:46

Saad

20.4k92352

asked Mar 22 at 0:33

mushimastermushimaster

16510

$endgroup$

closed as off-topic by RRL, Saad, Leucippus, Eevee Trainer, Wouter Mar 22 at 9:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Leucippus, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How come you don't have any examples?
  $endgroup$
  – Minus One-Twelfth
  Mar 22 at 0:35
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I'm guessing that you should have a limit somewhere in you question, right?
  $endgroup$
  – D.B.
  Mar 22 at 0:35
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You need to finish your sentence!
  $endgroup$
  – Ted Shifrin
  Mar 22 at 0:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you show that $frac1n - 1 to 0$ using $varepsilon$s and $N$s?
  $endgroup$
  – Theo Bendit
  Mar 22 at 0:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Saad, it is always better to type the Greek name of Greek letters (such as epsilon) rather than paste them as characters, because the latter version belongs to font families that maybe not all browsers can display (depending on what fonts a user has installed). Second, people talk about you, and I myself believe that yout behaviour regarding edits is wrong.
  $endgroup$
  – Alex M.
  Mar 23 at 21:38
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Prove that, for any $ninmathbbN$,$$fracn^22n^3-1$$using an $ε$-$N$ proof.

I am suppose to prove the sequence above converge using $ε$-$N$ proof. However there is no example in my textbook, I am suppose to use what I know about $ε$-$N$ proofs, choosing an appropriate $ε$.

sequences-and-series

share|cite|improve this question

edited Mar 22 at 2:46

Saad

20.4k92352

asked Mar 22 at 0:33

mushimastermushimaster

16510

$endgroup$

share|cite|improve this question

edited Mar 22 at 2:46

Saad

20.4k92352

asked Mar 22 at 0:33

mushimastermushimaster

16510

share|cite|improve this question

edited Mar 22 at 2:46

Saad

20.4k92352

asked Mar 22 at 0:33

mushimastermushimaster

16510

edited Mar 22 at 2:46

Saad

20.4k92352

edited Mar 22 at 2:46

Saad

20.4k92352

asked Mar 22 at 0:33

mushimastermushimaster

16510

asked Mar 22 at 0:33

mushimastermushimaster

16510

2 Answers
2

active

oldest

votes

1

$begingroup$

We write

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3; tag 1$

for $n > 2$,

$n^-3 < dfrac18, tag 2$

whence

$2 - n^-3 > dfrac158, tag 3$

$dfrac12 - n^-3 < dfrac815, tag 4$

$dfracn^-12 - n^-3 < dfrac815 n^-1; tag 5$

now for any $epsilon > 0$, we may choose $N_0$ sufficiently large that

$n > N_0 Longrightarrow dfrac815 n^-1 < epsilon; tag 6$

indeed, with

$N_0 > dfrac815 epsilon, tag 7$

we have

$dfrac815 epsilon < N_0 < n Longrightarrow dfrac1epsilon < dfrac158 n Longrightarrow dfrac815 n^-1 < epsilon; tag 8$

combining (1), (5), and (8) yields

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3 < dfrac815 n^-1 < epsilon tag 9$

for $n > N_0$; since we may choose $epsilon$ arbitrarily small, and $N_0$ correspondingly large, we have shown that

$displaystyle lim_n to infty dfracn^22n^3 - 1 = 0, tag10$

the desired result.

share|cite|improve this answer

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

$endgroup$

add a comment | 

0

$begingroup$

You have

• 
  $0 leq fracn^22n^3-1 < fracn^2n^3-n^2=frac1n-1$ for $n > 1$
  


• For $epsilon > 0$ you have $frac1n-1 < epsilon Leftrightarrow n > frac1epsilon+1$
  

Setting $N_epsilon = lfloor frac1epsilon+1rfloor + 1$ you have $forall n geq N_epsilon$:
$$left|fracn^22n^3-1 - 0 right| = fracn^22n^3-1 < frac1n-1 < epsilon$$
This shows using the $epsilon-N$ way that
$$lim_nto inftyfracn^22n^3-1 = 0$$

share|cite|improve this answer

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

1

$begingroup$

We write

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3; tag 1$

for $n > 2$,

$n^-3 < dfrac18, tag 2$

whence

$2 - n^-3 > dfrac158, tag 3$

$dfrac12 - n^-3 < dfrac815, tag 4$

$dfracn^-12 - n^-3 < dfrac815 n^-1; tag 5$

now for any $epsilon > 0$, we may choose $N_0$ sufficiently large that

$n > N_0 Longrightarrow dfrac815 n^-1 < epsilon; tag 6$

indeed, with

$N_0 > dfrac815 epsilon, tag 7$

we have

$dfrac815 epsilon < N_0 < n Longrightarrow dfrac1epsilon < dfrac158 n Longrightarrow dfrac815 n^-1 < epsilon; tag 8$

combining (1), (5), and (8) yields

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3 < dfrac815 n^-1 < epsilon tag 9$

for $n > N_0$; since we may choose $epsilon$ arbitrarily small, and $N_0$ correspondingly large, we have shown that

$displaystyle lim_n to infty dfracn^22n^3 - 1 = 0, tag10$

the desired result.

share|cite|improve this answer

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

$endgroup$

add a comment | 

1

$begingroup$

We write

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3; tag 1$

for $n > 2$,

$n^-3 < dfrac18, tag 2$

whence

$2 - n^-3 > dfrac158, tag 3$

$dfrac12 - n^-3 < dfrac815, tag 4$

$dfracn^-12 - n^-3 < dfrac815 n^-1; tag 5$

now for any $epsilon > 0$, we may choose $N_0$ sufficiently large that

$n > N_0 Longrightarrow dfrac815 n^-1 < epsilon; tag 6$

indeed, with

$N_0 > dfrac815 epsilon, tag 7$

we have

$dfrac815 epsilon < N_0 < n Longrightarrow dfrac1epsilon < dfrac158 n Longrightarrow dfrac815 n^-1 < epsilon; tag 8$

combining (1), (5), and (8) yields

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3 < dfrac815 n^-1 < epsilon tag 9$

for $n > N_0$; since we may choose $epsilon$ arbitrarily small, and $N_0$ correspondingly large, we have shown that

$displaystyle lim_n to infty dfracn^22n^3 - 1 = 0, tag10$

the desired result.

share|cite|improve this answer

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

$endgroup$

add a comment | 

$begingroup$

We write

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3; tag 1$

for $n > 2$,

$n^-3 < dfrac18, tag 2$

whence

$2 - n^-3 > dfrac158, tag 3$

$dfrac12 - n^-3 < dfrac815, tag 4$

$dfracn^-12 - n^-3 < dfrac815 n^-1; tag 5$

now for any $epsilon > 0$, we may choose $N_0$ sufficiently large that

$n > N_0 Longrightarrow dfrac815 n^-1 < epsilon; tag 6$

indeed, with

$N_0 > dfrac815 epsilon, tag 7$

we have

$dfrac815 epsilon < N_0 < n Longrightarrow dfrac1epsilon < dfrac158 n Longrightarrow dfrac815 n^-1 < epsilon; tag 8$

combining (1), (5), and (8) yields

$dfracn^22n^3 - 1 = dfracn^-12 - n^-3 < dfrac815 n^-1 < epsilon tag 9$

for $n > N_0$; since we may choose $epsilon$ arbitrarily small, and $N_0$ correspondingly large, we have shown that

$displaystyle lim_n to infty dfracn^22n^3 - 1 = 0, tag10$

the desired result.

share|cite|improve this answer

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

$endgroup$

share|cite|improve this answer

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

answered Mar 22 at 2:28

Robert LewisRobert Lewis

48.7k23167

0

$begingroup$

You have

• 
  $0 leq fracn^22n^3-1 < fracn^2n^3-n^2=frac1n-1$ for $n > 1$
  


• For $epsilon > 0$ you have $frac1n-1 < epsilon Leftrightarrow n > frac1epsilon+1$
  

Setting $N_epsilon = lfloor frac1epsilon+1rfloor + 1$ you have $forall n geq N_epsilon$:
$$left|fracn^22n^3-1 - 0 right| = fracn^22n^3-1 < frac1n-1 < epsilon$$
This shows using the $epsilon-N$ way that
$$lim_nto inftyfracn^22n^3-1 = 0$$

share|cite|improve this answer

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

0

$begingroup$

You have

• 
  $0 leq fracn^22n^3-1 < fracn^2n^3-n^2=frac1n-1$ for $n > 1$
  


• For $epsilon > 0$ you have $frac1n-1 < epsilon Leftrightarrow n > frac1epsilon+1$
  

Setting $N_epsilon = lfloor frac1epsilon+1rfloor + 1$ you have $forall n geq N_epsilon$:
$$left|fracn^22n^3-1 - 0 right| = fracn^22n^3-1 < frac1n-1 < epsilon$$
This shows using the $epsilon-N$ way that
$$lim_nto inftyfracn^22n^3-1 = 0$$

share|cite|improve this answer

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

$endgroup$

add a comment | 

$begingroup$

You have

• 
  $0 leq fracn^22n^3-1 < fracn^2n^3-n^2=frac1n-1$ for $n > 1$
  


• For $epsilon > 0$ you have $frac1n-1 < epsilon Leftrightarrow n > frac1epsilon+1$
  

Setting $N_epsilon = lfloor frac1epsilon+1rfloor + 1$ you have $forall n geq N_epsilon$:
$$left|fracn^22n^3-1 - 0 right| = fracn^22n^3-1 < frac1n-1 < epsilon$$
This shows using the $epsilon-N$ way that
$$lim_nto inftyfracn^22n^3-1 = 0$$

share|cite|improve this answer

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

$endgroup$

share|cite|improve this answer

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828

answered Mar 22 at 7:21

trancelocationtrancelocation

13.6k1828