Does $sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}}$ converge?
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I have $$sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}}$$
First I check whether it converges absolutely (in which case in converges).
$$
sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}} = sum_{n=1}^infty {frac{1}{sqrt n}} times |e^{i{frac{pi}{n}}}| = sum_{n=1}^infty {frac{1}{sqrt n}} times 1
$$
Since ${frac{1}{sqrt n}}$ converges then I'd say the whole series converges.
But the answer is that it doesn't converge.
What am I doing wrong?
sequences-and-series complex-analysis convergence
asked Jan 21 at 17:51
user3132457user3132457
1598
$endgroup$
add a comment |
$begingroup$
I have $$sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}}$$
First I check whether it converges absolutely (in which case in converges).
$$
sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}} = sum_{n=1}^infty {frac{1}{sqrt n}} times |e^{i{frac{pi}{n}}}| = sum_{n=1}^infty {frac{1}{sqrt n}} times 1
$$
Since ${frac{1}{sqrt n}}$ converges then I'd say the whole series converges.
But the answer is that it doesn't converge.
What am I doing wrong?
sequences-and-series complex-analysis convergence
asked Jan 21 at 17:51
user3132457user3132457
1598
$endgroup$
2
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
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How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
2
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56
add a comment |
$begingroup$
I have $$sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}}$$
First I check whether it converges absolutely (in which case in converges).
$$
sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}} = sum_{n=1}^infty {frac{1}{sqrt n}} times |e^{i{frac{pi}{n}}}| = sum_{n=1}^infty {frac{1}{sqrt n}} times 1
$$
Since ${frac{1}{sqrt n}}$ converges then I'd say the whole series converges.
But the answer is that it doesn't converge.
What am I doing wrong?
sequences-and-series complex-analysis convergence
asked Jan 21 at 17:51
user3132457user3132457
1598
$endgroup$
I have $$sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}}$$
First I check whether it converges absolutely (in which case in converges).
$$
sum_{n=1}^infty {frac{e^{i{frac{pi}{n}}}}{sqrt n}} = sum_{n=1}^infty {frac{1}{sqrt n}} times |e^{i{frac{pi}{n}}}| = sum_{n=1}^infty {frac{1}{sqrt n}} times 1
$$
Since ${frac{1}{sqrt n}}$ converges then I'd say the whole series converges.
But the answer is that it doesn't converge.
What am I doing wrong?
sequences-and-series complex-analysis convergence
sequences-and-series complex-analysis convergence
asked Jan 21 at 17:51
user3132457user3132457
1598
asked Jan 21 at 17:51
user3132457user3132457
1598
asked Jan 21 at 17:51
user3132457user3132457
1598
asked Jan 21 at 17:51
user3132457user3132457
1598
asked Jan 21 at 17:51
user3132457user3132457
1598
1598
2
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
$begingroup$
How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
2
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56
add a comment |
2
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
$begingroup$
How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
2
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56
2
2
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
$begingroup$
How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
$begingroup$
How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
2
2
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56
add a comment |
1 Answer
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$begingroup$
The series $sum_{n=1}^inftyfrac1{sqrt n}$ diverges. Therefore, your argument is flawed.
On the other hand$$(forall ninmathbb{N}):operatorname{Re}left(frac{e^{ifracpi n}}{sqrt n}right)=frac{cosleft(fracpi nright)}{sqrt n}geqslantfrac1{2sqrt n}$$if $n$ is large enough. Therefore, yes, your series diverges.
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
The series $sum_{n=1}^inftyfrac1{sqrt n}$ diverges. Therefore, your argument is flawed.
On the other hand$$(forall ninmathbb{N}):operatorname{Re}left(frac{e^{ifracpi n}}{sqrt n}right)=frac{cosleft(fracpi nright)}{sqrt n}geqslantfrac1{2sqrt n}$$if $n$ is large enough. Therefore, yes, your series diverges.
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
The series $sum_{n=1}^inftyfrac1{sqrt n}$ diverges. Therefore, your argument is flawed.
On the other hand$$(forall ninmathbb{N}):operatorname{Re}left(frac{e^{ifracpi n}}{sqrt n}right)=frac{cosleft(fracpi nright)}{sqrt n}geqslantfrac1{2sqrt n}$$if $n$ is large enough. Therefore, yes, your series diverges.
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
The series $sum_{n=1}^inftyfrac1{sqrt n}$ diverges. Therefore, your argument is flawed.
On the other hand$$(forall ninmathbb{N}):operatorname{Re}left(frac{e^{ifracpi n}}{sqrt n}right)=frac{cosleft(fracpi nright)}{sqrt n}geqslantfrac1{2sqrt n}$$if $n$ is large enough. Therefore, yes, your series diverges.
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
The series $sum_{n=1}^inftyfrac1{sqrt n}$ diverges. Therefore, your argument is flawed.
On the other hand$$(forall ninmathbb{N}):operatorname{Re}left(frac{e^{ifracpi n}}{sqrt n}right)=frac{cosleft(fracpi nright)}{sqrt n}geqslantfrac1{2sqrt n}$$if $n$ is large enough. Therefore, yes, your series diverges.
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 21 at 17:56
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
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2
$begingroup$
$1/sqrt{n}$ doesn't converge.
$endgroup$
– Wojowu
Jan 21 at 17:54
$begingroup$
How would you prove that the series of $n^{-1/2}$ converges? Any way, if your series comverged, then its real part would; however the real part is equivalent to $n^{-1/2} >0$ thus it will diverge.
$endgroup$
– Mindlack
Jan 21 at 17:55
2
$begingroup$
$1/sqrt{n}$ converges, but the relevant thing is $sum 1/sqrt{n}$ diverges.
$endgroup$
– GEdgar
Jan 21 at 17:56