Finding $limlimits_{n→∞}frac{ln(n)}nsumlimits_{k=1}^nfrac{a_k}k$
$begingroup$
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?
real-analysis
edited Dec 23 '18 at 3:54
Saad
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
$endgroup$
add a comment |
$begingroup$
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?
real-analysis
edited Dec 23 '18 at 3:54
Saad
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
$endgroup$
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
add a comment |
$begingroup$
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?
real-analysis
edited Dec 23 '18 at 3:54
Saad
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
$endgroup$
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
real-analysis
edited Dec 23 '18 at 3:54
Saad
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
edited Dec 23 '18 at 3:54
Saad
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
edited Dec 23 '18 at 3:54
Saad
20.3k92352
edited Dec 23 '18 at 3:54
Saad
20.3k92352
edited Dec 23 '18 at 3:54
Saad
20.3k92352
20.3k92352
asked Dec 23 '18 at 2:31
mavericmaveric
89712
asked Dec 23 '18 at 2:31
mavericmaveric
89712
asked Dec 23 '18 at 2:31
mavericmaveric
89712
89712
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
$endgroup$
$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
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
$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
$endgroup$
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
$$lim_{n rightarrow infty}frac{ln(n)}{n}sum_{k=1}^nfrac {a_k}{k}=0$$
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
answered Dec 23 '18 at 3:42
glowstonetreesglowstonetrees
2,380418
2,380418
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
$endgroup$
$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04
add a comment |
$begingroup$
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 $0$.
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
$endgroup$
$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04
add a comment |
$begingroup$
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 $0$.
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
$endgroup$
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 $0$.
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
answered Dec 23 '18 at 2:40
Ben MillwoodBen Millwood
11.4k32049
11.4k32049
$begingroup$
could not understand the last few lines.
$endgroup$
– maveric
Dec 23 '18 at 3:04
add a comment |
$begingroup$
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
add a comment |
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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