Mixing
Uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$
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What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?
The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.
real-analysis sequences-and-series uniform-convergence
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
$endgroup$
add a comment |
$begingroup$
What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?
The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.
real-analysis sequences-and-series uniform-convergence
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
$endgroup$
$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38
add a comment |
$begingroup$
What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?
The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.
real-analysis sequences-and-series uniform-convergence
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
$endgroup$
What can be said about the uniform Convergence of $sum_{n=1}^{infty}frac{x}{[(n-1)x+1][nx+1]}$ in the interval $[0,1]$?
The sequence inside the summation bracket doesn't seem to yield to root or ratio tests. The pointwise convergence itself seems doubtful. Should we use Cauchy-Criterion directly here? Maybe some comparison would be useful in this case? And what about uniform convergence? Any hints? Thanks beforehand.
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
edited Jan 24 at 12:57
José Carlos Santos
168k22132236
168k22132236
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
asked Jan 24 at 12:49
vidyarthividyarthi
3,0341833
3,0341833
$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38
add a comment |
$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38
$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38
add a comment |
2 Answers
2
active
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Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
add a comment |
$begingroup$
The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
add a comment |
$begingroup$
Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
add a comment |
$begingroup$
Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
Hint: $displaystylefrac1{bigl((n-1)x+1bigr)(nx+1)}=frac1{(n-1)x+1}-frac1{nx+1}$.
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
answered Jan 24 at 12:54
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
add a comment |
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
I just missed it though!
$endgroup$
– vidyarthi
Jan 24 at 13:00
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
though this would be biased, but I think my own answer above is more detailed, so you miss $25$ rep!
$endgroup$
– vidyarthi
Jan 25 at 9:58
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
$begingroup$
yes, done it and you got it
$endgroup$
– vidyarthi
Jan 25 at 10:01
add a comment |
$begingroup$
The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
$endgroup$
add a comment |
$begingroup$
The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
$endgroup$
add a comment |
$begingroup$
The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
$endgroup$
The series can be computed by using telescoping. We have $frac{x}{[(n-1)x+1][nx+1]}=frac1{(n-1)x+1}-frac1{nx+1}$. Thus, the sequence of partial sums would be $s_n=frac{nx}{nx+1}$. Hence, sum is equal to $$begin{cases}0 text{when} x=0\1 text{when} xneq0end{cases}$$. This easily shows non-uniform convergence, as the limit is discontinuous at $0$.
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
answered Jan 24 at 12:59
vidyarthividyarthi
3,0341833
3,0341833
add a comment |
add a comment |
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$begingroup$
Uniform convergence where?
$endgroup$
– José Carlos Santos
Jan 24 at 12:53
$begingroup$
@JoséCarlosSantos thanks! edited the question
$endgroup$
– vidyarthi
Jan 24 at 12:54
$begingroup$
What is allowed? Why don't you just compute partial sums directly?
$endgroup$
– metamorphy
Jan 25 at 7:38