Prove the complex series converge absolutely
$begingroup$
Prove the complex $$sum_{k=0}^infty (k+k^2i)^{-1}$$
series converge absolutely
Solution: by triangle inequality, we have
$$|k+k^2i|ge k^2-k ge {k^2over 2}$$ if $$k ge 2 $$
how to understand this??
sequences-and-series complex-analysis
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
$endgroup$
add a comment |
$begingroup$
Prove the complex $$sum_{k=0}^infty (k+k^2i)^{-1}$$
series converge absolutely
Solution: by triangle inequality, we have
$$|k+k^2i|ge k^2-k ge {k^2over 2}$$ if $$k ge 2 $$
how to understand this??
sequences-and-series complex-analysis
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
$endgroup$
$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26
add a comment |
$begingroup$
Prove the complex $$sum_{k=0}^infty (k+k^2i)^{-1}$$
series converge absolutely
Solution: by triangle inequality, we have
$$|k+k^2i|ge k^2-k ge {k^2over 2}$$ if $$k ge 2 $$
how to understand this??
sequences-and-series complex-analysis
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
$endgroup$
Prove the complex $$sum_{k=0}^infty (k+k^2i)^{-1}$$
series converge absolutely
Solution: by triangle inequality, we have
$$|k+k^2i|ge k^2-k ge {k^2over 2}$$ if $$k ge 2 $$
how to understand this??
sequences-and-series complex-analysis
sequences-and-series complex-analysis
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
edited Dec 9 '18 at 23:39
José Carlos Santos
156k22125227
156k22125227
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
asked Dec 9 '18 at 23:23
Ziang XingZiang Xing
61
61
$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26
add a comment |
$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26
$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26
$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am assuming that the sum begins with $k=1$.
What you did is correct and proves that the series converges absolutely. You can also use the fact that$$frac1{k+k^2i}=frac{k-k^2i}{k^2+k^k}=frac1{k+k^3}-frac1{1+k^2}i$$and that both series$$sum_{k=1}^inftyfrac1{k+k^3}text{ and }sum_{k=1}^inftyfrac1{1+k^2}$$converge, by the comparaison test.
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am assuming that the sum begins with $k=1$.
What you did is correct and proves that the series converges absolutely. You can also use the fact that$$frac1{k+k^2i}=frac{k-k^2i}{k^2+k^k}=frac1{k+k^3}-frac1{1+k^2}i$$and that both series$$sum_{k=1}^inftyfrac1{k+k^3}text{ and }sum_{k=1}^inftyfrac1{1+k^2}$$converge, by the comparaison test.
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
add a comment |
$begingroup$
I am assuming that the sum begins with $k=1$.
What you did is correct and proves that the series converges absolutely. You can also use the fact that$$frac1{k+k^2i}=frac{k-k^2i}{k^2+k^k}=frac1{k+k^3}-frac1{1+k^2}i$$and that both series$$sum_{k=1}^inftyfrac1{k+k^3}text{ and }sum_{k=1}^inftyfrac1{1+k^2}$$converge, by the comparaison test.
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
add a comment |
$begingroup$
I am assuming that the sum begins with $k=1$.
What you did is correct and proves that the series converges absolutely. You can also use the fact that$$frac1{k+k^2i}=frac{k-k^2i}{k^2+k^k}=frac1{k+k^3}-frac1{1+k^2}i$$and that both series$$sum_{k=1}^inftyfrac1{k+k^3}text{ and }sum_{k=1}^inftyfrac1{1+k^2}$$converge, by the comparaison test.
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
I am assuming that the sum begins with $k=1$.
What you did is correct and proves that the series converges absolutely. You can also use the fact that$$frac1{k+k^2i}=frac{k-k^2i}{k^2+k^k}=frac1{k+k^3}-frac1{1+k^2}i$$and that both series$$sum_{k=1}^inftyfrac1{k+k^3}text{ and }sum_{k=1}^inftyfrac1{1+k^2}$$converge, by the comparaison test.
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
answered Dec 9 '18 at 23:35
José Carlos SantosJosé Carlos Santos
156k22125227
156k22125227
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
add a comment |
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
how to understand that triangle inequality??
$endgroup$
– Ziang Xing
Dec 10 '18 at 0:09
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
$begingroup$
If $z,winmathbb C$, then $|z|=|z-w+w|leqslant|z-w|+|w|$ and therefore $|z-w|geqslant|z|-|w|$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:12
add a comment |
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$begingroup$
Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality?
$endgroup$
– GenericMathematician
Dec 9 '18 at 23:26