A non cauchy-schwarz approach to the question: Prove that if $sum{a_k^2}$ converges then $sum{a_k/k}$...












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 14:35







pahad2000

















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
















  • 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












2 Answers
2






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 19 at 7:10

























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


















  • $begingroup$
    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











1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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