Investigate convergence series $ sum_ {n = 1}^{infty} a^{frac {1} {sqrt [n] {2} -1}} $ for $ a in mathbb R $
$begingroup$
I know that this series is convergent for $a in <0,1)$ and divergent for $age1$, but I have a trouble with $a<0$ because my friend saw that for this $a$ it is not specified. Hovewer I don't know if he's right and why, because he can not explain it and I did not find it in any books. Can anyone dispel my doubts?
real-analysis sequences-and-series
asked Jan 13 at 14:57
MP3129MP3129
2237
$endgroup$
add a comment |
$begingroup$
I know that this series is convergent for $a in <0,1)$ and divergent for $age1$, but I have a trouble with $a<0$ because my friend saw that for this $a$ it is not specified. Hovewer I don't know if he's right and why, because he can not explain it and I did not find it in any books. Can anyone dispel my doubts?
real-analysis sequences-and-series
asked Jan 13 at 14:57
MP3129MP3129
2237
$endgroup$
1
$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
1
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18
add a comment |
$begingroup$
I know that this series is convergent for $a in <0,1)$ and divergent for $age1$, but I have a trouble with $a<0$ because my friend saw that for this $a$ it is not specified. Hovewer I don't know if he's right and why, because he can not explain it and I did not find it in any books. Can anyone dispel my doubts?
real-analysis sequences-and-series
asked Jan 13 at 14:57
MP3129MP3129
2237
$endgroup$
I know that this series is convergent for $a in <0,1)$ and divergent for $age1$, but I have a trouble with $a<0$ because my friend saw that for this $a$ it is not specified. Hovewer I don't know if he's right and why, because he can not explain it and I did not find it in any books. Can anyone dispel my doubts?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 13 at 14:57
MP3129MP3129
2237
asked Jan 13 at 14:57
MP3129MP3129
2237
edited Jan 13 at 15:23
MP3129
asked Jan 13 at 14:57
MP3129MP3129
2237
asked Jan 13 at 14:57
MP3129MP3129
2237
asked Jan 13 at 14:57
MP3129MP3129
2237
2237
1
$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
1
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18
add a comment |
1
$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
1
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18
1
1
$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
1
1
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18
add a comment |
1 Answer
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$begingroup$
For positive $a$, your series converges for $ain[0,1)$ You can see this because, using some Taylor expansions, $frac{1}{sqrt[n]{2}-1}sim frac{n}{log 2}-frac{1}{2}$. By the comparison test, your series's convergence is reduced to that of the geometric series. Thus, $a^{1/log(2)}<1, ainmathbb{R}^+ implies ain[0,1)$.
For negative $a$, use the fact that $big|sum_{nge 1}b_nbig|le sum_{nge 1} |b_n |$, which will show you that your series also converges for $ain(-1,0]$.
Thus, your series converge for $ain(-1,1)$.
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
$endgroup$
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
add a comment |
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$begingroup$
For positive $a$, your series converges for $ain[0,1)$ You can see this because, using some Taylor expansions, $frac{1}{sqrt[n]{2}-1}sim frac{n}{log 2}-frac{1}{2}$. By the comparison test, your series's convergence is reduced to that of the geometric series. Thus, $a^{1/log(2)}<1, ainmathbb{R}^+ implies ain[0,1)$.
For negative $a$, use the fact that $big|sum_{nge 1}b_nbig|le sum_{nge 1} |b_n |$, which will show you that your series also converges for $ain(-1,0]$.
Thus, your series converge for $ain(-1,1)$.
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
$endgroup$
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
add a comment |
$begingroup$
For positive $a$, your series converges for $ain[0,1)$ You can see this because, using some Taylor expansions, $frac{1}{sqrt[n]{2}-1}sim frac{n}{log 2}-frac{1}{2}$. By the comparison test, your series's convergence is reduced to that of the geometric series. Thus, $a^{1/log(2)}<1, ainmathbb{R}^+ implies ain[0,1)$.
For negative $a$, use the fact that $big|sum_{nge 1}b_nbig|le sum_{nge 1} |b_n |$, which will show you that your series also converges for $ain(-1,0]$.
Thus, your series converge for $ain(-1,1)$.
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
$endgroup$
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
add a comment |
$begingroup$
For positive $a$, your series converges for $ain[0,1)$ You can see this because, using some Taylor expansions, $frac{1}{sqrt[n]{2}-1}sim frac{n}{log 2}-frac{1}{2}$. By the comparison test, your series's convergence is reduced to that of the geometric series. Thus, $a^{1/log(2)}<1, ainmathbb{R}^+ implies ain[0,1)$.
For negative $a$, use the fact that $big|sum_{nge 1}b_nbig|le sum_{nge 1} |b_n |$, which will show you that your series also converges for $ain(-1,0]$.
Thus, your series converge for $ain(-1,1)$.
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
$endgroup$
For positive $a$, your series converges for $ain[0,1)$ You can see this because, using some Taylor expansions, $frac{1}{sqrt[n]{2}-1}sim frac{n}{log 2}-frac{1}{2}$. By the comparison test, your series's convergence is reduced to that of the geometric series. Thus, $a^{1/log(2)}<1, ainmathbb{R}^+ implies ain[0,1)$.
For negative $a$, use the fact that $big|sum_{nge 1}b_nbig|le sum_{nge 1} |b_n |$, which will show you that your series also converges for $ain(-1,0]$.
Thus, your series converge for $ain(-1,1)$.
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
answered Jan 13 at 16:11
ZacharyZachary
2,3451214
2,3451214
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
add a comment |
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
You're taking for granted that the series is even defined for negative $a$. Since the powers are irrational, this may not be true in the context of a real analysis class.
$endgroup$
– Ben W
Jan 13 at 16:13
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
$begingroup$
Yes, the series may not be real, but it is still perfectly reasonable to investigate its convergence. As long as the series of the norms of the terms converge, the series will also converge. I would also argue that even if this question is posed in the context of a real analysis class, it is fine to acknowledge that the sequence may not be real for $a<0$ and deal with its norms. I am not using any strictly complex analytic mathematical tools here; all of the techniques in my proof are commonly taught to real analysis students—who should not completely ignore the existence of complex numbers.
$endgroup$
– Zachary
Jan 13 at 16:24
add a comment |
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$begingroup$
I think you have some typos in your question statement. The series converges for $0leq a<1$ and diverges for $1leq a<infty$. For the case where $a<0$ I am not sure the series is defined, unless there is some weird definition involving complex numbers. In that case analogous facts would probably hold. But I think it is just undefined for $a<0$.
$endgroup$
– Ben W
Jan 13 at 15:18
$begingroup$
@BenW yes, you have right that I had some typos and I corrected it. So also you do not know the claim that speaks of this indeterminacy :/ In that case, maybe you have some idea how to justify it, if I had this question on the exam?
$endgroup$
– MP3129
Jan 13 at 15:25
1
$begingroup$
The problem is that your exponents are irrational. In the context of real analysis, irrational exponentiation with base $a$ is only defined for $0leq a<infty$. In the context of complex analysis exponentiation can be defined for complex $a$ (and in particular for $a<0$). However, this usually involves a branch cut of the log function, and this is clearly beyond the scope of a real analysis course. See here for more info: en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers
$endgroup$
– Ben W
Jan 13 at 16:18