A non cauchy-schwarz approach to the question: Prove that if $sum{a_k^2}$ converges then $sum{a_k/k}$...
$begingroup$
I wanted to know how to solve this question without the use of the cauchy-schwarz inequality and using more standard methods/tests (such as ratio test, comparison test etc.) for convergence/divergence if possible. Thanks in advance.
sequences-and-series convergence
asked Jan 17 at 13:41
pahad2000pahad2000
164
$endgroup$
add a comment |
$begingroup$
I wanted to know how to solve this question without the use of the cauchy-schwarz inequality and using more standard methods/tests (such as ratio test, comparison test etc.) for convergence/divergence if possible. Thanks in advance.
sequences-and-series convergence
asked Jan 17 at 13:41
pahad2000pahad2000
164
$endgroup$
1
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
2
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
3
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18
add a comment |
$begingroup$
I wanted to know how to solve this question without the use of the cauchy-schwarz inequality and using more standard methods/tests (such as ratio test, comparison test etc.) for convergence/divergence if possible. Thanks in advance.
sequences-and-series convergence
asked Jan 17 at 13:41
pahad2000pahad2000
164
$endgroup$
I wanted to know how to solve this question without the use of the cauchy-schwarz inequality and using more standard methods/tests (such as ratio test, comparison test etc.) for convergence/divergence if possible. Thanks in advance.
sequences-and-series convergence
sequences-and-series convergence
asked Jan 17 at 13:41
pahad2000pahad2000
164
asked Jan 17 at 13:41
pahad2000pahad2000
164
edited Jan 17 at 14:35
pahad2000
asked Jan 17 at 13:41
pahad2000pahad2000
164
asked Jan 17 at 13:41
pahad2000pahad2000
164
asked Jan 17 at 13:41
pahad2000pahad2000
164
164
1
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
2
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
3
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18
add a comment |
1
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
2
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
3
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18
1
1
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
2
2
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
3
3
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: (assuming all $a_k in mathbb{R}$)
Just note that according to Cauchy-Schwarz inequality you have
$$left|sum_{k=1}^N frac{a_k}{k} right| leq sqrt{sum_{k=1}^Nfrac{1}{k^2}}cdot sqrt{sum_{k=1}^N a_k^2}$$
Edit after OP changed the question to "non-Cauchy-Schwarz":
You may use the inequality GM-QM (geometric/quadratic mean)
or you derive directly that
- $|a_kcdotfrac{1}{k}| leq frac{a_k^2 + frac{1}{k^2}}{2}$
$$Rightarrow left|sum_{k=1}^N frac{a_k}{k} right| leq sum_{k=1}^N left|frac{a_k}{k} right| leq frac{1}{2}left( sum_{k=1}^Na_k^2+ sum_{k=1}^N frac{1}{k^2} right)$$
Now the required result follows by comparison.
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
add a comment |
$begingroup$
Using Cauchy–Schwarz inequality, you have
$$left(sum_{k=1}^n frac{a_k}{k} right)^2 le sum_{k=1}^n a_k^2 sum_{k=1}^n 1/k^2$$
and both series on the right side converge.
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
Hint: (assuming all $a_k in mathbb{R}$)
Just note that according to Cauchy-Schwarz inequality you have
$$left|sum_{k=1}^N frac{a_k}{k} right| leq sqrt{sum_{k=1}^Nfrac{1}{k^2}}cdot sqrt{sum_{k=1}^N a_k^2}$$
Edit after OP changed the question to "non-Cauchy-Schwarz":
You may use the inequality GM-QM (geometric/quadratic mean)
or you derive directly that
- $|a_kcdotfrac{1}{k}| leq frac{a_k^2 + frac{1}{k^2}}{2}$
$$Rightarrow left|sum_{k=1}^N frac{a_k}{k} right| leq sum_{k=1}^N left|frac{a_k}{k} right| leq frac{1}{2}left( sum_{k=1}^Na_k^2+ sum_{k=1}^N frac{1}{k^2} right)$$
Now the required result follows by comparison.
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
add a comment |
$begingroup$
Hint: (assuming all $a_k in mathbb{R}$)
Just note that according to Cauchy-Schwarz inequality you have
$$left|sum_{k=1}^N frac{a_k}{k} right| leq sqrt{sum_{k=1}^Nfrac{1}{k^2}}cdot sqrt{sum_{k=1}^N a_k^2}$$
Edit after OP changed the question to "non-Cauchy-Schwarz":
You may use the inequality GM-QM (geometric/quadratic mean)
or you derive directly that
- $|a_kcdotfrac{1}{k}| leq frac{a_k^2 + frac{1}{k^2}}{2}$
$$Rightarrow left|sum_{k=1}^N frac{a_k}{k} right| leq sum_{k=1}^N left|frac{a_k}{k} right| leq frac{1}{2}left( sum_{k=1}^Na_k^2+ sum_{k=1}^N frac{1}{k^2} right)$$
Now the required result follows by comparison.
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
add a comment |
$begingroup$
Hint: (assuming all $a_k in mathbb{R}$)
Just note that according to Cauchy-Schwarz inequality you have
$$left|sum_{k=1}^N frac{a_k}{k} right| leq sqrt{sum_{k=1}^Nfrac{1}{k^2}}cdot sqrt{sum_{k=1}^N a_k^2}$$
Edit after OP changed the question to "non-Cauchy-Schwarz":
You may use the inequality GM-QM (geometric/quadratic mean)
or you derive directly that
- $|a_kcdotfrac{1}{k}| leq frac{a_k^2 + frac{1}{k^2}}{2}$
$$Rightarrow left|sum_{k=1}^N frac{a_k}{k} right| leq sum_{k=1}^N left|frac{a_k}{k} right| leq frac{1}{2}left( sum_{k=1}^Na_k^2+ sum_{k=1}^N frac{1}{k^2} right)$$
Now the required result follows by comparison.
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
$endgroup$
Hint: (assuming all $a_k in mathbb{R}$)
Just note that according to Cauchy-Schwarz inequality you have
$$left|sum_{k=1}^N frac{a_k}{k} right| leq sqrt{sum_{k=1}^Nfrac{1}{k^2}}cdot sqrt{sum_{k=1}^N a_k^2}$$
Edit after OP changed the question to "non-Cauchy-Schwarz":
You may use the inequality GM-QM (geometric/quadratic mean)
or you derive directly that
- $|a_kcdotfrac{1}{k}| leq frac{a_k^2 + frac{1}{k^2}}{2}$
$$Rightarrow left|sum_{k=1}^N frac{a_k}{k} right| leq sum_{k=1}^N left|frac{a_k}{k} right| leq frac{1}{2}left( sum_{k=1}^Na_k^2+ sum_{k=1}^N frac{1}{k^2} right)$$
Now the required result follows by comparison.
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
edited Jan 19 at 7:10
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
answered Jan 17 at 13:45
trancelocationtrancelocation
14.1k1829
14.1k1829
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
add a comment |
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
Thanks. But is there an alternative solution you can think of perhaps just using the standard tests for convergence/divergence?
$endgroup$
– pahad2000
Jan 17 at 13:54
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
$begingroup$
There aren't restrictions on $a_k$ or $b_k$.
$endgroup$
– pahad2000
Jan 17 at 13:56
1
1
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
@pahad2000 I changed the solution according to your change of the question.
$endgroup$
– trancelocation
Jan 19 at 7:26
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
$begingroup$
Thanks so much!
$endgroup$
– pahad2000
Jan 22 at 16:34
add a comment |
$begingroup$
Using Cauchy–Schwarz inequality, you have
$$left(sum_{k=1}^n frac{a_k}{k} right)^2 le sum_{k=1}^n a_k^2 sum_{k=1}^n 1/k^2$$
and both series on the right side converge.
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
add a comment |
$begingroup$
Using Cauchy–Schwarz inequality, you have
$$left(sum_{k=1}^n frac{a_k}{k} right)^2 le sum_{k=1}^n a_k^2 sum_{k=1}^n 1/k^2$$
and both series on the right side converge.
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
add a comment |
$begingroup$
Using Cauchy–Schwarz inequality, you have
$$left(sum_{k=1}^n frac{a_k}{k} right)^2 le sum_{k=1}^n a_k^2 sum_{k=1}^n 1/k^2$$
and both series on the right side converge.
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
Using Cauchy–Schwarz inequality, you have
$$left(sum_{k=1}^n frac{a_k}{k} right)^2 le sum_{k=1}^n a_k^2 sum_{k=1}^n 1/k^2$$
and both series on the right side converge.
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 17 at 13:46
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
add a comment |
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
$begingroup$
Thank you. I'm a bit tentative to apply the cauchy-schwarz inequality. Could there be an alternative solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:57
add a comment |
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1
$begingroup$
use CS inequality
$endgroup$
– RAM_3R
Jan 17 at 13:47
2
$begingroup$
$able{1over 2}(a^2+b^2)$.
$endgroup$
– David Mitra
Jan 17 at 13:52
$begingroup$
Thanks so much. I haven't learn't the cauchy-schwarz inequality yet and so feel that there may be an alternative route through the problem. Is there a different solution that sticks to the standard tests for convergence/divergence only?
$endgroup$
– pahad2000
Jan 17 at 13:59
3
$begingroup$
Use the Comparison Test. $|a_k|(1/k)le (1/2)(a_k^2+ 1/k^2)$ from my previous comment (which is proved from $0le (a-b)^2$)..
$endgroup$
– David Mitra
Jan 17 at 15:18