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A Cautionary Suggestion

A sequence that has integer values for prime indexes only:

True or false? Then $x_n/y_n$ $rightarrow$ 0.

Show that $x_n rightarrow 0$Inequality for $|f(x_n)-f(x_n+(y_n-x_n)|$.Lim inf $ (x_n + y_n)$ less than or equal to lim inf $x_n +$ lim sup $y_n$Shouldn't $liminf(x_n)+liminf(y_n)=liminf(x_n+y_n)$ always hold true?True or false: problems on sequencesAs $left| x_n - y_n right|<frac1n$ and $x_n to pi$ then $y_nto pi$Let $x_n, y_n in mathbbR$, $lim x_n=a$, $lim y_n = b$ then $|x_n-b|<r<|y_n-a| forall n implies r = |a-b|$Convergence of $(x_n)$, $(x_n + y_n)$ and $(y_n)$Prove that if $x_n leq y_n, forall n$ then the same inequality holds for their limits, i.e. $xleq y$Let $z_n = x_n + y_n$, with $(x_n)$ and $(y_n)$ strictly increasing. Prove that if $(z_n)$ is bounded above, then so are $(x_n)$ and $(y_n)$.

0

$begingroup$

Suppose that $x_n$ is bounded, $y_n$>0 for all n, and $y_n$ $rightarrow$ + $infty$

True or false? Then $x_n/y_n$ $rightarrow$ 0.

This is what I have:

Given $bigx_nbig$ is bounded, let $epsilon>$0 there $exists$ $M>0$, M$in$ $mathbbN$ st $|x_n|leq$ M $forall$ n $in$ $mathbbN$. Then, $fracepsilonM$>0 and $exists$ $M_0$ $in$ $mathbbN$ st $|frac1y_n|<$ $fracepsilonM$ for all n $geq$ $M_0$. Then $|fracx_ny_n-0|$= $|fracx_ny_n|$ $leq$ $fracM$ =$fracM$ < M* $fracepsilonM$. Divide out the M and $fracM$ < $epsilon$ when n $geq$ $m_0$. Therefore, $|fracx_ny_n-0|$ < $epsilon$ $forall$ n $geq$ $m_0$. So, $lim_ntoinfty$ $fracx_ny_n$ =0 meaning $fracx_ny_n$ $rightarrow$ 0. Therefore, the statement is true.

Would this be correct or do I have errors? if I have errors how do I fix it?

analysis

share|cite|improve this question

edited Mar 11 at 21:05

user597188

asked Mar 11 at 21:01

user597188user597188

234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Huh? What are the hypotheses?
  $endgroup$
  – Randall
  Mar 11 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OP this is written quite unclearly.
  $endgroup$
  – Mike
  Mar 11 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you don't tell us what xn or yn are
  $endgroup$
  – user29418
  Mar 11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is fixed... sorry
  $endgroup$
  – user597188
  Mar 11 at 21:05
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose that $x_n$ is bounded, $y_n$>0 for all n, and $y_n$ $rightarrow$ + $infty$

True or false? Then $x_n/y_n$ $rightarrow$ 0.

This is what I have:

Given $bigx_nbig$ is bounded, let $epsilon>$0 there $exists$ $M>0$, M$in$ $mathbbN$ st $|x_n|leq$ M $forall$ n $in$ $mathbbN$. Then, $fracepsilonM$>0 and $exists$ $M_0$ $in$ $mathbbN$ st $|frac1y_n|<$ $fracepsilonM$ for all n $geq$ $M_0$. Then $|fracx_ny_n-0|$= $|fracx_ny_n|$ $leq$ $fracM$ =$fracM$ < M* $fracepsilonM$. Divide out the M and $fracM$ < $epsilon$ when n $geq$ $m_0$. Therefore, $|fracx_ny_n-0|$ < $epsilon$ $forall$ n $geq$ $m_0$. So, $lim_ntoinfty$ $fracx_ny_n$ =0 meaning $fracx_ny_n$ $rightarrow$ 0. Therefore, the statement is true.

Would this be correct or do I have errors? if I have errors how do I fix it?

analysis

share|cite|improve this question

edited Mar 11 at 21:05

user597188

asked Mar 11 at 21:01

user597188user597188

234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Huh? What are the hypotheses?
  $endgroup$
  – Randall
  Mar 11 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OP this is written quite unclearly.
  $endgroup$
  – Mike
  Mar 11 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you don't tell us what xn or yn are
  $endgroup$
  – user29418
  Mar 11 at 21:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is fixed... sorry
  $endgroup$
  – user597188
  Mar 11 at 21:05
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that $x_n$ is bounded, $y_n$>0 for all n, and $y_n$ $rightarrow$ + $infty$

True or false? Then $x_n/y_n$ $rightarrow$ 0.

This is what I have:

Given $bigx_nbig$ is bounded, let $epsilon>$0 there $exists$ $M>0$, M$in$ $mathbbN$ st $|x_n|leq$ M $forall$ n $in$ $mathbbN$. Then, $fracepsilonM$>0 and $exists$ $M_0$ $in$ $mathbbN$ st $|frac1y_n|<$ $fracepsilonM$ for all n $geq$ $M_0$. Then $|fracx_ny_n-0|$= $|fracx_ny_n|$ $leq$ $fracM$ =$fracM$ < M* $fracepsilonM$. Divide out the M and $fracM$ < $epsilon$ when n $geq$ $m_0$. Therefore, $|fracx_ny_n-0|$ < $epsilon$ $forall$ n $geq$ $m_0$. So, $lim_ntoinfty$ $fracx_ny_n$ =0 meaning $fracx_ny_n$ $rightarrow$ 0. Therefore, the statement is true.

Would this be correct or do I have errors? if I have errors how do I fix it?

analysis

share|cite|improve this question

edited Mar 11 at 21:05

user597188

asked Mar 11 at 21:01

user597188user597188

234

$endgroup$

share|cite|improve this question

edited Mar 11 at 21:05

user597188

asked Mar 11 at 21:01

user597188user597188

234

share|cite|improve this question

edited Mar 11 at 21:05

user597188

asked Mar 11 at 21:01

user597188user597188

234

asked Mar 11 at 21:01

user597188user597188

234

asked Mar 11 at 21:01

user597188user597188

234