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Let $a_n $ complex sequence prove that if $ a_nto infty$ then $|a_n|toinfty$. Note that $a_n = x_n + y_ni$

Limits of series proofs help neededIf $(a_n)$ is a complex sequence such that $sum_n=1^inftyfraca_nk^n = 0,forall kin mathbbN$ then $a_n = 0$ for all $ n$Let $a_n$ be a sequence such that: $ a_n+1-a_n ge frac1n$. prove that $lim_limitsn to infty a_n = infty$.Prove that if $(a_n)$ is a monotone increasing sequence of integers then $lim_ntoinfty(1+frac1a_n)^a_n=e$$(a_n)$ is a monotone increasing sequence of integers. Prove that: $lim_ntoinfty(1+frac1a_n)^a_n=e$Prove that $limsup_n to infty x_n cdot y_n = x cdotlimsup_n to infty y_n$Prove that if $lim limits_n to infty$ $x_n$ = $L$, then $lim limits_n to infty$ $|x_n|$ = $|L|$.Prove that every sequence $a_n$, $n in mathbbN, a_nneq 0$, that converges to $0$ satisfies the following:Prove that $lim_nto inftyS_n = infty$Prove that if $lim_ntoinfty x_n = 1$ for $x_n > 0$ then $lim_ntoinfty sqrt[n]x_1x_2cdots x_n = 1$

0

$begingroup$

Let $a_n $ complex sequence prove that if $ a_nto infty$ then $|a_n|toinfty$.
Note that $a_n = x_n + y_ni$
i dont know how to write that mathmatically.

trial :

Can i say that for every $M>0$ there exist $N$ such that for every $n>N$ ,

$~~|x_n|>M~~ OR
~~~|y_n|>M$ ( At least one of them goes to $infty$)

because of that $|an| = sqrt(x_n)^2+(y_n)^2 > M$ and so $|a_n|toinfty$.

complex-analysis limits complex-numbers

share|cite|improve this question

asked Mar 17 at 21:29

Mather Mather

4028

$endgroup$

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  $begingroup$
  What is your definition of $a_n to infty$ for complex $a_n$? I am asking because I would take your assertion as a definition for $a_n to infty$.
  $endgroup$
  – Klaus
  Mar 17 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont know thats what is confusing me , can i see your definition it might help
  $endgroup$
  – Mather
  Mar 17 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would you prove this i think the question is simple
  $endgroup$
  – Mather
  Mar 17 at 21:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would say $a_n to infty :Longleftrightarrow |a_n| to infty$, but that is what you want to prove, so I don't know what you're asking.
  $endgroup$
  – Klaus
  Mar 17 at 21:35
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $a_n $ complex sequence prove that if $ a_nto infty$ then $|a_n|toinfty$.
Note that $a_n = x_n + y_ni$
i dont know how to write that mathmatically.

trial :

Can i say that for every $M>0$ there exist $N$ such that for every $n>N$ ,

$~~|x_n|>M~~ OR
~~~|y_n|>M$ ( At least one of them goes to $infty$)

because of that $|an| = sqrt(x_n)^2+(y_n)^2 > M$ and so $|a_n|toinfty$.

complex-analysis limits complex-numbers

share|cite|improve this question

asked Mar 17 at 21:29

Mather Mather

4028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your definition of $a_n to infty$ for complex $a_n$? I am asking because I would take your assertion as a definition for $a_n to infty$.
  $endgroup$
  – Klaus
  Mar 17 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont know thats what is confusing me , can i see your definition it might help
  $endgroup$
  – Mather
  Mar 17 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how would you prove this i think the question is simple
  $endgroup$
  – Mather
  Mar 17 at 21:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would say $a_n to infty :Longleftrightarrow |a_n| to infty$, but that is what you want to prove, so I don't know what you're asking.
  $endgroup$
  – Klaus
  Mar 17 at 21:35
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $a_n $ complex sequence prove that if $ a_nto infty$ then $|a_n|toinfty$.
Note that $a_n = x_n + y_ni$
i dont know how to write that mathmatically.

trial :

Can i say that for every $M>0$ there exist $N$ such that for every $n>N$ ,

$~~|x_n|>M~~ OR
~~~|y_n|>M$ ( At least one of them goes to $infty$)

because of that $|an| = sqrt(x_n)^2+(y_n)^2 > M$ and so $|a_n|toinfty$.

complex-analysis limits complex-numbers

share|cite|improve this question

asked Mar 17 at 21:29

Mather Mather

4028

$endgroup$

share|cite|improve this question

asked Mar 17 at 21:29

Mather Mather

4028

share|cite|improve this question

asked Mar 17 at 21:29

Mather Mather

4028

asked Mar 17 at 21:29

Mather Mather

4028

asked Mar 17 at 21:29

Mather Mather

4028