Comparison Test for complex series: logical argument?












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$begingroup$


I am trying to show that the comparison test holds for complex series, meaning: if $sum_{n=0}^{infty} z_n $ is a complex series and $sum_{n=0}^{infty} a_n $ is a convergent non-negative real number series and $|z_n| leq a_n forall n in mathbb{N} $ then $sum_{n=0}^{infty} z_n $ converges.



To do so, I wanted to show that $|z_n - z_m| leq |a_n - a| + |a_m - a| < epsilon$ to get a Cauchy convergent series. My question is, is it logical my argument as it follows?



We have:



$$ |z_n - z_m|^2 = |z_n|^2 + |z_m|^2 - (z_noverline{z_m}) - (overline{z_n}z_m) = |z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) leq (|z_n| - |z_m|)^2 $$ because $Re(z) leq |z| $. Hence:



$$|z_n - z_m| leq |z_n| - |z_m| $$



and since $|z_n| geq 0, forall n in mathbb{N}$, it is obvious that $ |z_n| - |z_m| leq |z_n|+|z_m| $.



Is my argument logical or am I missing a point?



Thank you people so much for your help! :)






sequences-and-series complex-analysis convergence complex-numbers cauchy-sequences







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  • $begingroup$
    Which number is $a$?
    $endgroup$
    – José Carlos Santos
    Dec 30 '18 at 17:50
















0












$begingroup$


I am trying to show that the comparison test holds for complex series, meaning: if $sum_{n=0}^{infty} z_n $ is a complex series and $sum_{n=0}^{infty} a_n $ is a convergent non-negative real number series and $|z_n| leq a_n forall n in mathbb{N} $ then $sum_{n=0}^{infty} z_n $ converges.



To do so, I wanted to show that $|z_n - z_m| leq |a_n - a| + |a_m - a| < epsilon$ to get a Cauchy convergent series. My question is, is it logical my argument as it follows?



We have:



$$ |z_n - z_m|^2 = |z_n|^2 + |z_m|^2 - (z_noverline{z_m}) - (overline{z_n}z_m) = |z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) leq (|z_n| - |z_m|)^2 $$ because $Re(z) leq |z| $. Hence:



$$|z_n - z_m| leq |z_n| - |z_m| $$



and since $|z_n| geq 0, forall n in mathbb{N}$, it is obvious that $ |z_n| - |z_m| leq |z_n|+|z_m| $.



Is my argument logical or am I missing a point?



Thank you people so much for your help! :)






sequences-and-series complex-analysis convergence complex-numbers cauchy-sequences







share|cite|improve this question









$endgroup$












  • $begingroup$
    Which number is $a$?
    $endgroup$
    – José Carlos Santos
    Dec 30 '18 at 17:50














0












0








0





$begingroup$


I am trying to show that the comparison test holds for complex series, meaning: if $sum_{n=0}^{infty} z_n $ is a complex series and $sum_{n=0}^{infty} a_n $ is a convergent non-negative real number series and $|z_n| leq a_n forall n in mathbb{N} $ then $sum_{n=0}^{infty} z_n $ converges.



To do so, I wanted to show that $|z_n - z_m| leq |a_n - a| + |a_m - a| < epsilon$ to get a Cauchy convergent series. My question is, is it logical my argument as it follows?



We have:



$$ |z_n - z_m|^2 = |z_n|^2 + |z_m|^2 - (z_noverline{z_m}) - (overline{z_n}z_m) = |z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) leq (|z_n| - |z_m|)^2 $$ because $Re(z) leq |z| $. Hence:



$$|z_n - z_m| leq |z_n| - |z_m| $$



and since $|z_n| geq 0, forall n in mathbb{N}$, it is obvious that $ |z_n| - |z_m| leq |z_n|+|z_m| $.



Is my argument logical or am I missing a point?



Thank you people so much for your help! :)






sequences-and-series complex-analysis convergence complex-numbers cauchy-sequences







share|cite|improve this question









$endgroup$




I am trying to show that the comparison test holds for complex series, meaning: if $sum_{n=0}^{infty} z_n $ is a complex series and $sum_{n=0}^{infty} a_n $ is a convergent non-negative real number series and $|z_n| leq a_n forall n in mathbb{N} $ then $sum_{n=0}^{infty} z_n $ converges.



To do so, I wanted to show that $|z_n - z_m| leq |a_n - a| + |a_m - a| < epsilon$ to get a Cauchy convergent series. My question is, is it logical my argument as it follows?



We have:



$$ |z_n - z_m|^2 = |z_n|^2 + |z_m|^2 - (z_noverline{z_m}) - (overline{z_n}z_m) = |z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) leq (|z_n| - |z_m|)^2 $$ because $Re(z) leq |z| $. Hence:



$$|z_n - z_m| leq |z_n| - |z_m| $$



and since $|z_n| geq 0, forall n in mathbb{N}$, it is obvious that $ |z_n| - |z_m| leq |z_n|+|z_m| $.



Is my argument logical or am I missing a point?



Thank you people so much for your help! :)






sequences-and-series complex-analysis convergence complex-numbers cauchy-sequences




sequences-and-series complex-analysis convergence complex-numbers cauchy-sequences






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asked Dec 30 '18 at 17:32









M.GonzalezM.Gonzalez

1126




1126












  • $begingroup$
    Which number is $a$?
    $endgroup$
    – José Carlos Santos
    Dec 30 '18 at 17:50


















  • $begingroup$
    Which number is $a$?
    $endgroup$
    – José Carlos Santos
    Dec 30 '18 at 17:50
















$begingroup$
Which number is $a$?
$endgroup$
– José Carlos Santos
Dec 30 '18 at 17:50




$begingroup$
Which number is $a$?
$endgroup$
– José Carlos Santos
Dec 30 '18 at 17:50










1 Answer
1






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1












$begingroup$

I think you are confusing a few things:



First, you want to show that $sum_{n=0}^infty z_n$ converges, but you check whether $(z_n)_n$ is a Cauchy sequence, which is something different (If $z_n=1$ for all $n$ then $(z_n)_n$ is a Cauchy sequence but the series is divergent).



Second, you have the inequality $|z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) stackrel{??}{leq} (|z_n| - |z_m|)^2 $ backwards (the LHS is greater than or equal to the RHS), since from $operatorname{Re} z leq |z|$ you get $-operatorname{Re} z geq -|z|$.



I would do this proof by looking at the partial sums $Z_N = sum_{n=0}^N z_n$ and $A_N = sum_{n=0}^N a_n$ and use the fact that $(A_N)_N$ is a Cauchy sequence to prove that $(Z_N)_N$ is a Cauchy sequence.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
    $endgroup$
    – M.Gonzalez
    Dec 31 '18 at 0:46











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1 Answer
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1 Answer
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active

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1












$begingroup$

I think you are confusing a few things:



First, you want to show that $sum_{n=0}^infty z_n$ converges, but you check whether $(z_n)_n$ is a Cauchy sequence, which is something different (If $z_n=1$ for all $n$ then $(z_n)_n$ is a Cauchy sequence but the series is divergent).



Second, you have the inequality $|z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) stackrel{??}{leq} (|z_n| - |z_m|)^2 $ backwards (the LHS is greater than or equal to the RHS), since from $operatorname{Re} z leq |z|$ you get $-operatorname{Re} z geq -|z|$.



I would do this proof by looking at the partial sums $Z_N = sum_{n=0}^N z_n$ and $A_N = sum_{n=0}^N a_n$ and use the fact that $(A_N)_N$ is a Cauchy sequence to prove that $(Z_N)_N$ is a Cauchy sequence.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
    $endgroup$
    – M.Gonzalez
    Dec 31 '18 at 0:46
















1












$begingroup$

I think you are confusing a few things:



First, you want to show that $sum_{n=0}^infty z_n$ converges, but you check whether $(z_n)_n$ is a Cauchy sequence, which is something different (If $z_n=1$ for all $n$ then $(z_n)_n$ is a Cauchy sequence but the series is divergent).



Second, you have the inequality $|z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) stackrel{??}{leq} (|z_n| - |z_m|)^2 $ backwards (the LHS is greater than or equal to the RHS), since from $operatorname{Re} z leq |z|$ you get $-operatorname{Re} z geq -|z|$.



I would do this proof by looking at the partial sums $Z_N = sum_{n=0}^N z_n$ and $A_N = sum_{n=0}^N a_n$ and use the fact that $(A_N)_N$ is a Cauchy sequence to prove that $(Z_N)_N$ is a Cauchy sequence.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
    $endgroup$
    – M.Gonzalez
    Dec 31 '18 at 0:46














1












1








1





$begingroup$

I think you are confusing a few things:



First, you want to show that $sum_{n=0}^infty z_n$ converges, but you check whether $(z_n)_n$ is a Cauchy sequence, which is something different (If $z_n=1$ for all $n$ then $(z_n)_n$ is a Cauchy sequence but the series is divergent).



Second, you have the inequality $|z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) stackrel{??}{leq} (|z_n| - |z_m|)^2 $ backwards (the LHS is greater than or equal to the RHS), since from $operatorname{Re} z leq |z|$ you get $-operatorname{Re} z geq -|z|$.



I would do this proof by looking at the partial sums $Z_N = sum_{n=0}^N z_n$ and $A_N = sum_{n=0}^N a_n$ and use the fact that $(A_N)_N$ is a Cauchy sequence to prove that $(Z_N)_N$ is a Cauchy sequence.






share|cite|improve this answer









$endgroup$



I think you are confusing a few things:



First, you want to show that $sum_{n=0}^infty z_n$ converges, but you check whether $(z_n)_n$ is a Cauchy sequence, which is something different (If $z_n=1$ for all $n$ then $(z_n)_n$ is a Cauchy sequence but the series is divergent).



Second, you have the inequality $|z_n|^2 + |z_m|^2 - 2Re(overline{z_n}z_m) stackrel{??}{leq} (|z_n| - |z_m|)^2 $ backwards (the LHS is greater than or equal to the RHS), since from $operatorname{Re} z leq |z|$ you get $-operatorname{Re} z geq -|z|$.



I would do this proof by looking at the partial sums $Z_N = sum_{n=0}^N z_n$ and $A_N = sum_{n=0}^N a_n$ and use the fact that $(A_N)_N$ is a Cauchy sequence to prove that $(Z_N)_N$ is a Cauchy sequence.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 18:10









0x5390x539

1,127317




1,127317












  • $begingroup$
    Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
    $endgroup$
    – M.Gonzalez
    Dec 31 '18 at 0:46


















  • $begingroup$
    Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
    $endgroup$
    – M.Gonzalez
    Dec 31 '18 at 0:46
















$begingroup$
Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
$endgroup$
– M.Gonzalez
Dec 31 '18 at 0:46




$begingroup$
Thank you, indeed LHS is greater or equal tk RHS! I found a solution from your tip! :)
$endgroup$
– M.Gonzalez
Dec 31 '18 at 0:46


















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