Finding $limlimits_{n→∞}frac{ln(n)}nsumlimits_{k=1}^nfrac{a_k}k$












-2












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For a given sequence $a_1,a_2,a_3,…,a_n$, if $limlimits_{n→∞}=a$, find$$lim_{n→∞}frac{ln(n)}nsum_{k=1}^nfrac{a_k}k.$$
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    $begingroup$
    It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
    $endgroup$
    – Josh B.
    Dec 23 '18 at 2:36






  • 1




    $begingroup$
    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    $endgroup$
    – Shaun
    Dec 23 '18 at 2:51
















-2












$begingroup$


For a given sequence $a_1,a_2,a_3,…,a_n$, if $limlimits_{n→∞}=a$, find$$lim_{n→∞}frac{ln(n)}nsum_{k=1}^nfrac{a_k}k.$$
Can there be no question which a person is not even able to start? If I have no idea how to start a question how can I say what I did?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
    $endgroup$
    – Josh B.
    Dec 23 '18 at 2:36






  • 1




    $begingroup$
    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    $endgroup$
    – Shaun
    Dec 23 '18 at 2:51














-2












-2








-2


0



$begingroup$


For a given sequence $a_1,a_2,a_3,…,a_n$, if $limlimits_{n→∞}=a$, find$$lim_{n→∞}frac{ln(n)}nsum_{k=1}^nfrac{a_k}k.$$
Can there be no question which a person is not even able to start? If I have no idea how to start a question how can I say what I did?










share|cite|improve this question











$endgroup$




For a given sequence $a_1,a_2,a_3,…,a_n$, if $limlimits_{n→∞}=a$, find$$lim_{n→∞}frac{ln(n)}nsum_{k=1}^nfrac{a_k}k.$$
Can there be no question which a person is not even able to start? If I have no idea how to start a question how can I say what I did?







real-analysis






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edited Dec 23 '18 at 3:54









Saad

19.7k92352




19.7k92352










asked Dec 23 '18 at 2:31









mavericmaveric

79112




79112








  • 1




    $begingroup$
    It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
    $endgroup$
    – Josh B.
    Dec 23 '18 at 2:36






  • 1




    $begingroup$
    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    $endgroup$
    – Shaun
    Dec 23 '18 at 2:51














  • 1




    $begingroup$
    It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
    $endgroup$
    – Josh B.
    Dec 23 '18 at 2:36






  • 1




    $begingroup$
    You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    $endgroup$
    – Shaun
    Dec 23 '18 at 2:51








1




1




$begingroup$
It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
$endgroup$
– Josh B.
Dec 23 '18 at 2:36




$begingroup$
It would be helpful if you worded your question a little more clearly. What is it you're trying to do?
$endgroup$
– Josh B.
Dec 23 '18 at 2:36




1




1




$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 23 '18 at 2:51




$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 23 '18 at 2:51










2 Answers
2






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votes


















2












$begingroup$

Let's start with what we know. We are given that $lim_{n rightarrow infty}a_n=a$, but you will see that we don't actually need to use this. All that matters is that the sequence is bounded (convergence implies bounded). i.e.



$$exists ; M in Bbb R ; | ; |a_k|<M ; ; forall ; k in Bbb N$$



Also, the Euler constant should come in handy, and it is defined as $$gamma := lim_{n rightarrow infty}bigg(sum_{k=1}^nfrac 1k - ln(n)bigg)$$



and implies that



$$lim_{n rightarrow infty}bigg|sum_{k=1}^nfrac 1k - ln(n)-gammabigg|=0$$



Using L'Hopital's rule, we see that



$$lim_{n rightarrow infty}frac{big[ln(n)big]^2}{n}=0 qquad text{ and } qquad lim_{n rightarrow infty}frac{ln(n)}{n}=0 $$



Thus, we have:



begin{align}
& bigg|frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k} bigg|\
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {|a_k|}{k}bigg| \
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {M}{k}bigg| \
= & Mfrac{ln(n)}{n}bigg(bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+bigg|ln(n)+gammabigg|bigg) \
= & Mbigg(frac{ln(n)}{n}bigg)bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+Mfrac{big[ln(n)big]^2}{n}+Mgamma frac{ln(n)}{n}\
rightarrow & M(0)|0| + M(0) + Mgamma (0) \
= & 0
end{align}



which implies that



$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$






share|cite|improve this answer









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  • $begingroup$
    will there be a different method which does not involve so advance concepts
    $endgroup$
    – maveric
    Dec 23 '18 at 9:42










  • $begingroup$
    which of the above concepts do you consider advanced?
    $endgroup$
    – glowstonetrees
    Dec 23 '18 at 16:03



















1












$begingroup$

First consider the case where $a_k = a$ for all $k$. Then the sum becomes a sum of $1/k$, which can be approximated by comparing it to the integral of $1/x$ in some way that I forget but involves the Euler-Mascheroni constant.



Anyway I think you get something that converges to $0$, and then you can argue that if $a_k to a$ then it isn't different enough from the above case, so will also converge to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    could not understand the last few lines.
    $endgroup$
    – maveric
    Dec 23 '18 at 3:04











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2 Answers
2






active

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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

Let's start with what we know. We are given that $lim_{n rightarrow infty}a_n=a$, but you will see that we don't actually need to use this. All that matters is that the sequence is bounded (convergence implies bounded). i.e.



$$exists ; M in Bbb R ; | ; |a_k|<M ; ; forall ; k in Bbb N$$



Also, the Euler constant should come in handy, and it is defined as $$gamma := lim_{n rightarrow infty}bigg(sum_{k=1}^nfrac 1k - ln(n)bigg)$$



and implies that



$$lim_{n rightarrow infty}bigg|sum_{k=1}^nfrac 1k - ln(n)-gammabigg|=0$$



Using L'Hopital's rule, we see that



$$lim_{n rightarrow infty}frac{big[ln(n)big]^2}{n}=0 qquad text{ and } qquad lim_{n rightarrow infty}frac{ln(n)}{n}=0 $$



Thus, we have:



begin{align}
& bigg|frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k} bigg|\
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {|a_k|}{k}bigg| \
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {M}{k}bigg| \
= & Mfrac{ln(n)}{n}bigg(bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+bigg|ln(n)+gammabigg|bigg) \
= & Mbigg(frac{ln(n)}{n}bigg)bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+Mfrac{big[ln(n)big]^2}{n}+Mgamma frac{ln(n)}{n}\
rightarrow & M(0)|0| + M(0) + Mgamma (0) \
= & 0
end{align}



which implies that



$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    will there be a different method which does not involve so advance concepts
    $endgroup$
    – maveric
    Dec 23 '18 at 9:42










  • $begingroup$
    which of the above concepts do you consider advanced?
    $endgroup$
    – glowstonetrees
    Dec 23 '18 at 16:03
















2












$begingroup$

Let's start with what we know. We are given that $lim_{n rightarrow infty}a_n=a$, but you will see that we don't actually need to use this. All that matters is that the sequence is bounded (convergence implies bounded). i.e.



$$exists ; M in Bbb R ; | ; |a_k|<M ; ; forall ; k in Bbb N$$



Also, the Euler constant should come in handy, and it is defined as $$gamma := lim_{n rightarrow infty}bigg(sum_{k=1}^nfrac 1k - ln(n)bigg)$$



and implies that



$$lim_{n rightarrow infty}bigg|sum_{k=1}^nfrac 1k - ln(n)-gammabigg|=0$$



Using L'Hopital's rule, we see that



$$lim_{n rightarrow infty}frac{big[ln(n)big]^2}{n}=0 qquad text{ and } qquad lim_{n rightarrow infty}frac{ln(n)}{n}=0 $$



Thus, we have:



begin{align}
& bigg|frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k} bigg|\
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {|a_k|}{k}bigg| \
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {M}{k}bigg| \
= & Mfrac{ln(n)}{n}bigg(bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+bigg|ln(n)+gammabigg|bigg) \
= & Mbigg(frac{ln(n)}{n}bigg)bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+Mfrac{big[ln(n)big]^2}{n}+Mgamma frac{ln(n)}{n}\
rightarrow & M(0)|0| + M(0) + Mgamma (0) \
= & 0
end{align}



which implies that



$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    will there be a different method which does not involve so advance concepts
    $endgroup$
    – maveric
    Dec 23 '18 at 9:42










  • $begingroup$
    which of the above concepts do you consider advanced?
    $endgroup$
    – glowstonetrees
    Dec 23 '18 at 16:03














2












2








2





$begingroup$

Let's start with what we know. We are given that $lim_{n rightarrow infty}a_n=a$, but you will see that we don't actually need to use this. All that matters is that the sequence is bounded (convergence implies bounded). i.e.



$$exists ; M in Bbb R ; | ; |a_k|<M ; ; forall ; k in Bbb N$$



Also, the Euler constant should come in handy, and it is defined as $$gamma := lim_{n rightarrow infty}bigg(sum_{k=1}^nfrac 1k - ln(n)bigg)$$



and implies that



$$lim_{n rightarrow infty}bigg|sum_{k=1}^nfrac 1k - ln(n)-gammabigg|=0$$



Using L'Hopital's rule, we see that



$$lim_{n rightarrow infty}frac{big[ln(n)big]^2}{n}=0 qquad text{ and } qquad lim_{n rightarrow infty}frac{ln(n)}{n}=0 $$



Thus, we have:



begin{align}
& bigg|frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k} bigg|\
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {|a_k|}{k}bigg| \
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {M}{k}bigg| \
= & Mfrac{ln(n)}{n}bigg(bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+bigg|ln(n)+gammabigg|bigg) \
= & Mbigg(frac{ln(n)}{n}bigg)bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+Mfrac{big[ln(n)big]^2}{n}+Mgamma frac{ln(n)}{n}\
rightarrow & M(0)|0| + M(0) + Mgamma (0) \
= & 0
end{align}



which implies that



$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$






share|cite|improve this answer









$endgroup$



Let's start with what we know. We are given that $lim_{n rightarrow infty}a_n=a$, but you will see that we don't actually need to use this. All that matters is that the sequence is bounded (convergence implies bounded). i.e.



$$exists ; M in Bbb R ; | ; |a_k|<M ; ; forall ; k in Bbb N$$



Also, the Euler constant should come in handy, and it is defined as $$gamma := lim_{n rightarrow infty}bigg(sum_{k=1}^nfrac 1k - ln(n)bigg)$$



and implies that



$$lim_{n rightarrow infty}bigg|sum_{k=1}^nfrac 1k - ln(n)-gammabigg|=0$$



Using L'Hopital's rule, we see that



$$lim_{n rightarrow infty}frac{big[ln(n)big]^2}{n}=0 qquad text{ and } qquad lim_{n rightarrow infty}frac{ln(n)}{n}=0 $$



Thus, we have:



begin{align}
& bigg|frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k} bigg|\
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {|a_k|}{k}bigg| \
leq & frac{ln(n)}{n}bigg|sum_{k=1}^nfrac {M}{k}bigg| \
= & Mfrac{ln(n)}{n}bigg(bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+bigg|ln(n)+gammabigg|bigg) \
= & Mbigg(frac{ln(n)}{n}bigg)bigg|sum_{k=1}^nfrac {1}{k}-ln(n)-gamma bigg|+Mfrac{big[ln(n)big]^2}{n}+Mgamma frac{ln(n)}{n}\
rightarrow & M(0)|0| + M(0) + Mgamma (0) \
= & 0
end{align}



which implies that



$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 3:42









glowstonetreesglowstonetrees

2,368418




2,368418












  • $begingroup$
    will there be a different method which does not involve so advance concepts
    $endgroup$
    – maveric
    Dec 23 '18 at 9:42










  • $begingroup$
    which of the above concepts do you consider advanced?
    $endgroup$
    – glowstonetrees
    Dec 23 '18 at 16:03


















  • $begingroup$
    will there be a different method which does not involve so advance concepts
    $endgroup$
    – maveric
    Dec 23 '18 at 9:42










  • $begingroup$
    which of the above concepts do you consider advanced?
    $endgroup$
    – glowstonetrees
    Dec 23 '18 at 16:03
















$begingroup$
will there be a different method which does not involve so advance concepts
$endgroup$
– maveric
Dec 23 '18 at 9:42




$begingroup$
will there be a different method which does not involve so advance concepts
$endgroup$
– maveric
Dec 23 '18 at 9:42












$begingroup$
which of the above concepts do you consider advanced?
$endgroup$
– glowstonetrees
Dec 23 '18 at 16:03




$begingroup$
which of the above concepts do you consider advanced?
$endgroup$
– glowstonetrees
Dec 23 '18 at 16:03











1












$begingroup$

First consider the case where $a_k = a$ for all $k$. Then the sum becomes a sum of $1/k$, which can be approximated by comparing it to the integral of $1/x$ in some way that I forget but involves the Euler-Mascheroni constant.



Anyway I think you get something that converges to $0$, and then you can argue that if $a_k to a$ then it isn't different enough from the above case, so will also converge to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    could not understand the last few lines.
    $endgroup$
    – maveric
    Dec 23 '18 at 3:04
















1












$begingroup$

First consider the case where $a_k = a$ for all $k$. Then the sum becomes a sum of $1/k$, which can be approximated by comparing it to the integral of $1/x$ in some way that I forget but involves the Euler-Mascheroni constant.



Anyway I think you get something that converges to $0$, and then you can argue that if $a_k to a$ then it isn't different enough from the above case, so will also converge to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    could not understand the last few lines.
    $endgroup$
    – maveric
    Dec 23 '18 at 3:04














1












1








1





$begingroup$

First consider the case where $a_k = a$ for all $k$. Then the sum becomes a sum of $1/k$, which can be approximated by comparing it to the integral of $1/x$ in some way that I forget but involves the Euler-Mascheroni constant.



Anyway I think you get something that converges to $0$, and then you can argue that if $a_k to a$ then it isn't different enough from the above case, so will also converge to $0$.






share|cite|improve this answer









$endgroup$



First consider the case where $a_k = a$ for all $k$. Then the sum becomes a sum of $1/k$, which can be approximated by comparing it to the integral of $1/x$ in some way that I forget but involves the Euler-Mascheroni constant.



Anyway I think you get something that converges to $0$, and then you can argue that if $a_k to a$ then it isn't different enough from the above case, so will also converge to $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 2:40









Ben MillwoodBen Millwood

11.3k32049




11.3k32049












  • $begingroup$
    could not understand the last few lines.
    $endgroup$
    – maveric
    Dec 23 '18 at 3:04


















  • $begingroup$
    could not understand the last few lines.
    $endgroup$
    – maveric
    Dec 23 '18 at 3:04
















$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04




$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04


















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