Is this sum convergent?Is this convergent sum a constant?Find sum of this convergent seriesIs series $sumlimits^infty_n=1fraccos(nx)n^alpha$, for $alpha>0$, convergent?A convergent series implies two split convergent seriesproof that $sum z_n$, $sum w_n$ convergent, then $sum (z_n+w_n)$ convergesIs the infinite sum convergent?Convergent series and its sumDecide if series convergent or divergent, and if convergent finding sum.Any information on this sumGiven that $sum b(n)$ is convergent, is $sum a(n)$ convergent?
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Is this sum convergent?
Is this convergent sum a constant?Find sum of this convergent seriesIs series $sumlimits^infty_n=1fraccos(nx)n^alpha$, for $alpha>0$, convergent?A convergent series implies two split convergent seriesproof that $sum z_n$, $sum w_n$ convergent, then $sum (z_n+w_n)$ convergesIs the infinite sum convergent?Convergent series and its sumDecide if series convergent or divergent, and if convergent finding sum.Any information on this sumGiven that $sum b(n)$ is convergent, is $sum a(n)$ convergent?
$begingroup$
The sum is: $$sum_3leq i,j,k < + infty frac 1ijk$$.
sequences-and-series convergence
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
The sum is: $$sum_3leq i,j,k < + infty frac 1ijk$$.
sequences-and-series convergence
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The sum is: $$sum_3leq i,j,k < + infty frac 1ijk$$.
sequences-and-series convergence
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The sum is: $$sum_3leq i,j,k < + infty frac 1ijk$$.
sequences-and-series convergence
sequences-and-series convergence
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Ante P.Ante P.
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Ante P.Ante P.
307
asked 8 hours ago
Ante P.Ante P.
307
307
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ante P. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your expression can be rewritten as
$$
left (sum_i=3^inftyfrac1iright )cdotleft (sum_j=3^inftyfrac1jright )cdotleft (sum_k=3^inftyfrac1kright )
=left (sum_i=3^inftyfrac1iright )^3.
$$
Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.
answered 8 hours ago
Marian G.Marian G.
41936
$endgroup$
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
add a comment |
$begingroup$
No, because it include the summands when $i=j=3,$ which are $$sum_k=3^inftyfrac19k,$$ which doesn't converge.
And if you want $3leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:
$$sum_k=5^inftyfrac112k,$$
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
$endgroup$
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
Clearly, $$sum_3le i,j,kfrac1ijk=sum_3le kfrac19k+sum_3le i-1,j-1,kfrac1ijk$$
$$>sum_kge3frac19k$$
and the Riemann series $sum frac 1k$ is divergent.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your expression can be rewritten as
$$
left (sum_i=3^inftyfrac1iright )cdotleft (sum_j=3^inftyfrac1jright )cdotleft (sum_k=3^inftyfrac1kright )
=left (sum_i=3^inftyfrac1iright )^3.
$$
Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.
answered 8 hours ago
Marian G.Marian G.
41936
$endgroup$
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
add a comment |
$begingroup$
Your expression can be rewritten as
$$
left (sum_i=3^inftyfrac1iright )cdotleft (sum_j=3^inftyfrac1jright )cdotleft (sum_k=3^inftyfrac1kright )
=left (sum_i=3^inftyfrac1iright )^3.
$$
Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.
answered 8 hours ago
Marian G.Marian G.
41936
$endgroup$
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
add a comment |
$begingroup$
Your expression can be rewritten as
$$
left (sum_i=3^inftyfrac1iright )cdotleft (sum_j=3^inftyfrac1jright )cdotleft (sum_k=3^inftyfrac1kright )
=left (sum_i=3^inftyfrac1iright )^3.
$$
Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.
answered 8 hours ago
Marian G.Marian G.
41936
$endgroup$
Your expression can be rewritten as
$$
left (sum_i=3^inftyfrac1iright )cdotleft (sum_j=3^inftyfrac1jright )cdotleft (sum_k=3^inftyfrac1kright )
=left (sum_i=3^inftyfrac1iright )^3.
$$
Since the last sum is divergent (e.g., by using the integral criterion), your original sum is also divergent.
answered 8 hours ago
Marian G.Marian G.
41936
answered 8 hours ago
Marian G.Marian G.
41936
answered 8 hours ago
Marian G.Marian G.
41936
answered 8 hours ago
Marian G.Marian G.
41936
41936
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
add a comment |
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
1
1
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
How to justify that transformation?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
$begingroup$
@AnteP. The expression on the left is the product of three the same infinite sums. The variable notation is unimportant. Therefore, the right-hand side is in the form of a cube.
$endgroup$
– Marian G.
8 hours ago
1
1
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
$begingroup$
I am asking how to justify the left hand side, not the right hand one.
$endgroup$
– Ante P.
8 hours ago
add a comment |
$begingroup$
No, because it include the summands when $i=j=3,$ which are $$sum_k=3^inftyfrac19k,$$ which doesn't converge.
And if you want $3leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:
$$sum_k=5^inftyfrac112k,$$
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
$endgroup$
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
No, because it include the summands when $i=j=3,$ which are $$sum_k=3^inftyfrac19k,$$ which doesn't converge.
And if you want $3leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:
$$sum_k=5^inftyfrac112k,$$
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
$endgroup$
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
No, because it include the summands when $i=j=3,$ which are $$sum_k=3^inftyfrac19k,$$ which doesn't converge.
And if you want $3leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:
$$sum_k=5^inftyfrac112k,$$
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
$endgroup$
No, because it include the summands when $i=j=3,$ which are $$sum_k=3^inftyfrac19k,$$ which doesn't converge.
And if you want $3leq i<j<k$ then you can take the cases when $i=3,j=4$ and get:
$$sum_k=5^inftyfrac112k,$$
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
edited 8 hours ago
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
answered 8 hours ago
Thomas AndrewsThomas Andrews
132k12149301
132k12149301
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Would anything change if we supposed that $i<j<k$?
$endgroup$
– Ante P.
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Nope. Added that to my answer. @AnteP.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
Clearly, $$sum_3le i,j,kfrac1ijk=sum_3le kfrac19k+sum_3le i-1,j-1,kfrac1ijk$$
$$>sum_kge3frac19k$$
and the Riemann series $sum frac 1k$ is divergent.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
$endgroup$
add a comment |
$begingroup$
Clearly, $$sum_3le i,j,kfrac1ijk=sum_3le kfrac19k+sum_3le i-1,j-1,kfrac1ijk$$
$$>sum_kge3frac19k$$
and the Riemann series $sum frac 1k$ is divergent.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
$endgroup$
add a comment |
$begingroup$
Clearly, $$sum_3le i,j,kfrac1ijk=sum_3le kfrac19k+sum_3le i-1,j-1,kfrac1ijk$$
$$>sum_kge3frac19k$$
and the Riemann series $sum frac 1k$ is divergent.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
$endgroup$
Clearly, $$sum_3le i,j,kfrac1ijk=sum_3le kfrac19k+sum_3le i-1,j-1,kfrac1ijk$$
$$>sum_kge3frac19k$$
and the Riemann series $sum frac 1k$ is divergent.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
edited 8 hours ago
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
38.4k21634
38.4k21634
add a comment |
add a comment |
Ante P. is a new contributor. Be nice, and check out our Code of Conduct.
Ante P. is a new contributor. Be nice, and check out our Code of Conduct.
Ante P. is a new contributor. Be nice, and check out our Code of Conduct.
Ante P. is a new contributor. Be nice, and check out our Code of Conduct.
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