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Let $sum_n=1^infty a_n$ for $a_ngt 0$. Show that $lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n$
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Let $sum_n=1^infty a_n$ be a series of positive terms. Show that $$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n$$
So this is what I could come up with. I hope it's right
$$ ta_1-1 lt lfloor ta_1rfloor le ta_1$$
$$ ta_2-1 lt lfloor ta_2rfloor le ta_2$$
$$vdots$$
$$ ta_n-1 lt lfloor ta_nrfloor le ta_n$$
$$vdots$$
Dividing with $t$ we get
$$a_1-frac 1t lt fraclfloor ta_1rfloort le a_1 $$
$$a_2-frac 1t lt fraclfloor ta_2rfloort le a_2 $$
$$vdots$$
$$a_n-frac 1t lt fraclfloor ta_nrfloort le a_n $$
$$vdots$$
If we use the limit when $t to infty$ in every inequality and then by summing all of them we get
$$sum_n=1^infty a_n le lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor le sum_n=1^infty a_n $$
So the limit is
$$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n $$
I'm not very sure if this is right. Can somebody tell me what to do or does someone has another method?
calculus sequences-and-series limits
asked Jul 27 at 12:15
J.Dane
159112
 |Â
show 2 more comments
up vote
2
down vote
favorite
Let $sum_n=1^infty a_n$ be a series of positive terms. Show that $$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n$$
So this is what I could come up with. I hope it's right
$$ ta_1-1 lt lfloor ta_1rfloor le ta_1$$
$$ ta_2-1 lt lfloor ta_2rfloor le ta_2$$
$$vdots$$
$$ ta_n-1 lt lfloor ta_nrfloor le ta_n$$
$$vdots$$
Dividing with $t$ we get
$$a_1-frac 1t lt fraclfloor ta_1rfloort le a_1 $$
$$a_2-frac 1t lt fraclfloor ta_2rfloort le a_2 $$
$$vdots$$
$$a_n-frac 1t lt fraclfloor ta_nrfloort le a_n $$
$$vdots$$
If we use the limit when $t to infty$ in every inequality and then by summing all of them we get
$$sum_n=1^infty a_n le lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor le sum_n=1^infty a_n $$
So the limit is
$$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n $$
I'm not very sure if this is right. Can somebody tell me what to do or does someone has another method?
calculus sequences-and-series limits
asked Jul 27 at 12:15
J.Dane
159112
There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
Is this right now?
– J.Dane
Jul 27 at 12:49
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $sum_n=1^infty a_n$ be a series of positive terms. Show that $$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n$$
So this is what I could come up with. I hope it's right
$$ ta_1-1 lt lfloor ta_1rfloor le ta_1$$
$$ ta_2-1 lt lfloor ta_2rfloor le ta_2$$
$$vdots$$
$$ ta_n-1 lt lfloor ta_nrfloor le ta_n$$
$$vdots$$
Dividing with $t$ we get
$$a_1-frac 1t lt fraclfloor ta_1rfloort le a_1 $$
$$a_2-frac 1t lt fraclfloor ta_2rfloort le a_2 $$
$$vdots$$
$$a_n-frac 1t lt fraclfloor ta_nrfloort le a_n $$
$$vdots$$
If we use the limit when $t to infty$ in every inequality and then by summing all of them we get
$$sum_n=1^infty a_n le lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor le sum_n=1^infty a_n $$
So the limit is
$$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n $$
I'm not very sure if this is right. Can somebody tell me what to do or does someone has another method?
calculus sequences-and-series limits
asked Jul 27 at 12:15
J.Dane
159112
Let $sum_n=1^infty a_n$ be a series of positive terms. Show that $$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n$$
So this is what I could come up with. I hope it's right
$$ ta_1-1 lt lfloor ta_1rfloor le ta_1$$
$$ ta_2-1 lt lfloor ta_2rfloor le ta_2$$
$$vdots$$
$$ ta_n-1 lt lfloor ta_nrfloor le ta_n$$
$$vdots$$
Dividing with $t$ we get
$$a_1-frac 1t lt fraclfloor ta_1rfloort le a_1 $$
$$a_2-frac 1t lt fraclfloor ta_2rfloort le a_2 $$
$$vdots$$
$$a_n-frac 1t lt fraclfloor ta_nrfloort le a_n $$
$$vdots$$
If we use the limit when $t to infty$ in every inequality and then by summing all of them we get
$$sum_n=1^infty a_n le lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor le sum_n=1^infty a_n $$
So the limit is
$$lim_tto inftyfrac1tsum_n=1^inftylfloor ta_nrfloor =sum_n=1^infty a_n $$
I'm not very sure if this is right. Can somebody tell me what to do or does someone has another method?
calculus sequences-and-series limits
asked Jul 27 at 12:15
J.Dane
159112
edited Jul 27 at 15:28
asked Jul 27 at 12:15
J.Dane
159112
asked Jul 27 at 12:15
J.Dane
159112
asked Jul 27 at 12:15
J.Dane
159112
159112
There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
Is this right now?
– J.Dane
Jul 27 at 12:49
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16
 |Â
show 2 more comments
There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
Is this right now?
– J.Dane
Jul 27 at 12:49
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16
There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
Is this right now?
– J.Dane
Jul 27 at 12:49
Is this right now?
– J.Dane
Jul 27 at 12:49
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
This should help: Let $S=sum_n=1^inftya_n$ and assume $S<infty.$ Let $epsilon>0.$ Then there exists $N$ such that $sum_n=1^Na_n > S-epsilon/2.$ Choose $t_0$ such that $N/t<epsilon/2$ for $tge t_0.$ Then $tge t_0$ implies
$$frac1tsum_n=1^inftylfloor ta_nrfloor ge frac1tsum_n=1^Nlfloor ta_n rfloor ge frac1tsum_n=1^N (ta_n-1) = sum_n=1^N a_n -N/t > S-epsilon/2 - epsilon/2 = S-epsilon.$$
answered Jul 27 at 16:02
zhw.
65.3k42870
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This should help: Let $S=sum_n=1^inftya_n$ and assume $S<infty.$ Let $epsilon>0.$ Then there exists $N$ such that $sum_n=1^Na_n > S-epsilon/2.$ Choose $t_0$ such that $N/t<epsilon/2$ for $tge t_0.$ Then $tge t_0$ implies
$$frac1tsum_n=1^inftylfloor ta_nrfloor ge frac1tsum_n=1^Nlfloor ta_n rfloor ge frac1tsum_n=1^N (ta_n-1) = sum_n=1^N a_n -N/t > S-epsilon/2 - epsilon/2 = S-epsilon.$$
answered Jul 27 at 16:02
zhw.
65.3k42870
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
add a comment |Â
up vote
1
down vote
accepted
This should help: Let $S=sum_n=1^inftya_n$ and assume $S<infty.$ Let $epsilon>0.$ Then there exists $N$ such that $sum_n=1^Na_n > S-epsilon/2.$ Choose $t_0$ such that $N/t<epsilon/2$ for $tge t_0.$ Then $tge t_0$ implies
$$frac1tsum_n=1^inftylfloor ta_nrfloor ge frac1tsum_n=1^Nlfloor ta_n rfloor ge frac1tsum_n=1^N (ta_n-1) = sum_n=1^N a_n -N/t > S-epsilon/2 - epsilon/2 = S-epsilon.$$
answered Jul 27 at 16:02
zhw.
65.3k42870
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This should help: Let $S=sum_n=1^inftya_n$ and assume $S<infty.$ Let $epsilon>0.$ Then there exists $N$ such that $sum_n=1^Na_n > S-epsilon/2.$ Choose $t_0$ such that $N/t<epsilon/2$ for $tge t_0.$ Then $tge t_0$ implies
$$frac1tsum_n=1^inftylfloor ta_nrfloor ge frac1tsum_n=1^Nlfloor ta_n rfloor ge frac1tsum_n=1^N (ta_n-1) = sum_n=1^N a_n -N/t > S-epsilon/2 - epsilon/2 = S-epsilon.$$
answered Jul 27 at 16:02
zhw.
65.3k42870
This should help: Let $S=sum_n=1^inftya_n$ and assume $S<infty.$ Let $epsilon>0.$ Then there exists $N$ such that $sum_n=1^Na_n > S-epsilon/2.$ Choose $t_0$ such that $N/t<epsilon/2$ for $tge t_0.$ Then $tge t_0$ implies
$$frac1tsum_n=1^inftylfloor ta_nrfloor ge frac1tsum_n=1^Nlfloor ta_n rfloor ge frac1tsum_n=1^N (ta_n-1) = sum_n=1^N a_n -N/t > S-epsilon/2 - epsilon/2 = S-epsilon.$$
answered Jul 27 at 16:02
zhw.
65.3k42870
answered Jul 27 at 16:02
zhw.
65.3k42870
answered Jul 27 at 16:02
zhw.
65.3k42870
answered Jul 27 at 16:02
zhw.
65.3k42870
65.3k42870
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
add a comment |Â
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Thank you very useful. So should I prove the other side of the inequality ie $frac1tsum_n=1^inftylfloor ta_nrfloor lefrac1tsum_n=1^infty ta_n= sum_n=1^infty a_n=S$ or is this enough?
– J.Dane
Jul 27 at 16:24
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Yes, that's the easy side. And you still have the $S=infty$ case to do.
– zhw.
Jul 27 at 16:25
Ok thank you again
– J.Dane
Jul 27 at 16:26
Ok thank you again
– J.Dane
Jul 27 at 16:26
add a comment |Â
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There are inequalities that you need to use such as $lfloor a_n rfloor lfloor t rfloor leq lfloor ta_n rfloor $
– Paul
Jul 27 at 12:28
Is this right now?
– J.Dane
Jul 27 at 12:49
What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient.
– JavaMan
Jul 27 at 13:04
We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms?
– J.Dane
Jul 27 at 13:11
Something is missing. When you sum $a_n-frac1t$ you end up with $sum_n=1^inftya_n-fracinftyt$. How did you get rid of $fracinftyt$, since the limit is $fracinftyinfty$?
– herb steinberg
Jul 27 at 15:16