Mixing
Finding value of a summation series.
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Find the value of
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k$$
Where $ine j,jne k,kne i$.
What does this notion means? How can we write this in expanded form? What does the inequality between $i,j,k$ mean? And most importantly how to solve this type of series(hints would be better)?
sequences-and-series summation
asked Aug 29 at 17:24
Love Invariants
90815
 |Â
show 6 more comments
up vote
3
down vote
favorite
Find the value of
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k$$
Where $ine j,jne k,kne i$.
What does this notion means? How can we write this in expanded form? What does the inequality between $i,j,k$ mean? And most importantly how to solve this type of series(hints would be better)?
sequences-and-series summation
asked Aug 29 at 17:24
Love Invariants
90815
You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36
 |Â
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the value of
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k$$
Where $ine j,jne k,kne i$.
What does this notion means? How can we write this in expanded form? What does the inequality between $i,j,k$ mean? And most importantly how to solve this type of series(hints would be better)?
sequences-and-series summation
asked Aug 29 at 17:24
Love Invariants
90815
Find the value of
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k$$
Where $ine j,jne k,kne i$.
What does this notion means? How can we write this in expanded form? What does the inequality between $i,j,k$ mean? And most importantly how to solve this type of series(hints would be better)?
sequences-and-series summation
asked Aug 29 at 17:24
Love Invariants
90815
edited Aug 29 at 17:37
asked Aug 29 at 17:24
Love Invariants
90815
asked Aug 29 at 17:24
Love Invariants
90815
asked Aug 29 at 17:24
Love Invariants
90815
90815
You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36
 |Â
show 6 more comments
You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36
You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36
 |Â
show 6 more comments
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
First ignore the restriction $i ne j ne k$. The result is $left( frac11 - frac13 right)^3 = frac278$, since this is simply the geometric series $sum_i=0^infty frac13^i$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $frac11 - frac13 cdot frac11 - frac19 = frac2716$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $frac11 - frac127 = frac2726$.
The result is
$$
frac278 - 3 cdotfrac2716 + 2 frac2726 approx 0.38942dots
$$
answered Aug 29 at 17:46
Hans Engler
9,55911836
add a comment |Â
up vote
1
down vote
Following the suggestion given in the comments we have that
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k
=left(sum^infty_k=01over3^kright)^3
-3sum^infty_k=01over3^ksum^infty_k=01over9^k
+2sum^infty_k=01over27^k$$
and by geometric series
$$sum^infty_k=01over3^k=frac11-frac13=frac32$$
$$sum^infty_k=01over9^k=frac11-frac19=frac98$$
$$sum^infty_k=01over27^k=frac11-frac127=frac2726$$
answered Aug 29 at 17:27
gimusi
70.8k73786
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
add a comment |Â
up vote
1
down vote
We can respect the conditions $ine j,jne k,kne i$ by adding up over $0leq i<j<k<infty$ and since the order counts we multiply with $3!$.
We obtain
beginalign*
colorblue3!&colorbluesum_0leq i<j<k<inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_k=j+1^inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_colorbluek=0^inftyfrac13^i+2j+k+1tag$kto k+j+1$\
&=3!sum_i=0^inftysum_colorbluej=0^inftysum_k=0^inftyfrac13^3i+2j+k+3tag$jto j+i+1$\
&=3!cdotfrac11-frac127cdotfrac11-frac19cdotfrac11-frac13cdotfrac127\
&,,colorblue=frac81208approx 0.38942
endalign*
answered Aug 29 at 20:53
Markus Scheuer
57k452136
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
First ignore the restriction $i ne j ne k$. The result is $left( frac11 - frac13 right)^3 = frac278$, since this is simply the geometric series $sum_i=0^infty frac13^i$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $frac11 - frac13 cdot frac11 - frac19 = frac2716$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $frac11 - frac127 = frac2726$.
The result is
$$
frac278 - 3 cdotfrac2716 + 2 frac2726 approx 0.38942dots
$$
answered Aug 29 at 17:46
Hans Engler
9,55911836
add a comment |Â
up vote
7
down vote
accepted
First ignore the restriction $i ne j ne k$. The result is $left( frac11 - frac13 right)^3 = frac278$, since this is simply the geometric series $sum_i=0^infty frac13^i$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $frac11 - frac13 cdot frac11 - frac19 = frac2716$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $frac11 - frac127 = frac2726$.
The result is
$$
frac278 - 3 cdotfrac2716 + 2 frac2726 approx 0.38942dots
$$
answered Aug 29 at 17:46
Hans Engler
9,55911836
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
First ignore the restriction $i ne j ne k$. The result is $left( frac11 - frac13 right)^3 = frac278$, since this is simply the geometric series $sum_i=0^infty frac13^i$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $frac11 - frac13 cdot frac11 - frac19 = frac2716$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $frac11 - frac127 = frac2726$.
The result is
$$
frac278 - 3 cdotfrac2716 + 2 frac2726 approx 0.38942dots
$$
answered Aug 29 at 17:46
Hans Engler
9,55911836
First ignore the restriction $i ne j ne k$. The result is $left( frac11 - frac13 right)^3 = frac278$, since this is simply the geometric series $sum_i=0^infty frac13^i$ raised to the power 3.
Now subtract the three series that come from the conditions $i=j, i=k, j=k$. These are each equal to $frac11 - frac13 cdot frac11 - frac19 = frac2716$.
Finally add back two times the series coming from the terms with $i = j = k$, since it has been subtracted three times and should only be subtracted once. This series is equal to $frac11 - frac127 = frac2726$.
The result is
$$
frac278 - 3 cdotfrac2716 + 2 frac2726 approx 0.38942dots
$$
answered Aug 29 at 17:46
Hans Engler
9,55911836
answered Aug 29 at 17:46
Hans Engler
9,55911836
answered Aug 29 at 17:46
Hans Engler
9,55911836
answered Aug 29 at 17:46
Hans Engler
9,55911836
9,55911836
add a comment |Â
add a comment |Â
up vote
1
down vote
Following the suggestion given in the comments we have that
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k
=left(sum^infty_k=01over3^kright)^3
-3sum^infty_k=01over3^ksum^infty_k=01over9^k
+2sum^infty_k=01over27^k$$
and by geometric series
$$sum^infty_k=01over3^k=frac11-frac13=frac32$$
$$sum^infty_k=01over9^k=frac11-frac19=frac98$$
$$sum^infty_k=01over27^k=frac11-frac127=frac2726$$
answered Aug 29 at 17:27
gimusi
70.8k73786
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
add a comment |Â
up vote
1
down vote
Following the suggestion given in the comments we have that
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k
=left(sum^infty_k=01over3^kright)^3
-3sum^infty_k=01over3^ksum^infty_k=01over9^k
+2sum^infty_k=01over27^k$$
and by geometric series
$$sum^infty_k=01over3^k=frac11-frac13=frac32$$
$$sum^infty_k=01over9^k=frac11-frac19=frac98$$
$$sum^infty_k=01over27^k=frac11-frac127=frac2726$$
answered Aug 29 at 17:27
gimusi
70.8k73786
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Following the suggestion given in the comments we have that
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k
=left(sum^infty_k=01over3^kright)^3
-3sum^infty_k=01over3^ksum^infty_k=01over9^k
+2sum^infty_k=01over27^k$$
and by geometric series
$$sum^infty_k=01over3^k=frac11-frac13=frac32$$
$$sum^infty_k=01over9^k=frac11-frac19=frac98$$
$$sum^infty_k=01over27^k=frac11-frac127=frac2726$$
answered Aug 29 at 17:27
gimusi
70.8k73786
Following the suggestion given in the comments we have that
$$sum^infty_i=0sum^infty_j=0sum^infty_k=01over3^i3^j3^k
=left(sum^infty_k=01over3^kright)^3
-3sum^infty_k=01over3^ksum^infty_k=01over9^k
+2sum^infty_k=01over27^k$$
and by geometric series
$$sum^infty_k=01over3^k=frac11-frac13=frac32$$
$$sum^infty_k=01over9^k=frac11-frac19=frac98$$
$$sum^infty_k=01over27^k=frac11-frac127=frac2726$$
answered Aug 29 at 17:27
gimusi
70.8k73786
edited Aug 29 at 17:53
answered Aug 29 at 17:27
gimusi
70.8k73786
answered Aug 29 at 17:27
gimusi
70.8k73786
answered Aug 29 at 17:27
gimusi
70.8k73786
70.8k73786
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
add a comment |Â
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
How does $ine jne k$ come into this?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@LordSharktheUnknown I thought it was simply a condition for independency of i,j,k. Is it wrong?
– gimusi
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
@gimusi-Yes, answer is 81/208
– Love Invariants
Aug 29 at 17:30
add a comment |Â
up vote
1
down vote
We can respect the conditions $ine j,jne k,kne i$ by adding up over $0leq i<j<k<infty$ and since the order counts we multiply with $3!$.
We obtain
beginalign*
colorblue3!&colorbluesum_0leq i<j<k<inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_k=j+1^inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_colorbluek=0^inftyfrac13^i+2j+k+1tag$kto k+j+1$\
&=3!sum_i=0^inftysum_colorbluej=0^inftysum_k=0^inftyfrac13^3i+2j+k+3tag$jto j+i+1$\
&=3!cdotfrac11-frac127cdotfrac11-frac19cdotfrac11-frac13cdotfrac127\
&,,colorblue=frac81208approx 0.38942
endalign*
answered Aug 29 at 20:53
Markus Scheuer
57k452136
add a comment |Â
up vote
1
down vote
We can respect the conditions $ine j,jne k,kne i$ by adding up over $0leq i<j<k<infty$ and since the order counts we multiply with $3!$.
We obtain
beginalign*
colorblue3!&colorbluesum_0leq i<j<k<inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_k=j+1^inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_colorbluek=0^inftyfrac13^i+2j+k+1tag$kto k+j+1$\
&=3!sum_i=0^inftysum_colorbluej=0^inftysum_k=0^inftyfrac13^3i+2j+k+3tag$jto j+i+1$\
&=3!cdotfrac11-frac127cdotfrac11-frac19cdotfrac11-frac13cdotfrac127\
&,,colorblue=frac81208approx 0.38942
endalign*
answered Aug 29 at 20:53
Markus Scheuer
57k452136
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We can respect the conditions $ine j,jne k,kne i$ by adding up over $0leq i<j<k<infty$ and since the order counts we multiply with $3!$.
We obtain
beginalign*
colorblue3!&colorbluesum_0leq i<j<k<inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_k=j+1^inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_colorbluek=0^inftyfrac13^i+2j+k+1tag$kto k+j+1$\
&=3!sum_i=0^inftysum_colorbluej=0^inftysum_k=0^inftyfrac13^3i+2j+k+3tag$jto j+i+1$\
&=3!cdotfrac11-frac127cdotfrac11-frac19cdotfrac11-frac13cdotfrac127\
&,,colorblue=frac81208approx 0.38942
endalign*
answered Aug 29 at 20:53
Markus Scheuer
57k452136
We can respect the conditions $ine j,jne k,kne i$ by adding up over $0leq i<j<k<infty$ and since the order counts we multiply with $3!$.
We obtain
beginalign*
colorblue3!&colorbluesum_0leq i<j<k<inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_k=j+1^inftyfrac13^i+j+k\
&=3!sum_i=0^inftysum_j=i+1^inftysum_colorbluek=0^inftyfrac13^i+2j+k+1tag$kto k+j+1$\
&=3!sum_i=0^inftysum_colorbluej=0^inftysum_k=0^inftyfrac13^3i+2j+k+3tag$jto j+i+1$\
&=3!cdotfrac11-frac127cdotfrac11-frac19cdotfrac11-frac13cdotfrac127\
&,,colorblue=frac81208approx 0.38942
endalign*
answered Aug 29 at 20:53
Markus Scheuer
57k452136
edited Aug 30 at 9:45
answered Aug 29 at 20:53
Markus Scheuer
57k452136
answered Aug 29 at 20:53
Markus Scheuer
57k452136
answered Aug 29 at 20:53
Markus Scheuer
57k452136
57k452136
add a comment |Â
add a comment |Â
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You allow $i=kne j$?
– Lord Shark the Unknown
Aug 29 at 17:28
@LordSharktheUnknown- No, thats a problem.
– Love Invariants
Aug 29 at 17:29
What does $ineq jneq k$ mean? Does it mean that in (i,j,k) there is no value occuring twice or three times?
– amsmath
Aug 29 at 17:29
If so, you should edit the question to make that clear.
– Lord Shark the Unknown
Aug 29 at 17:30
You should write $ineq j, jneq k, kneq i$, if you want to convey no two of $i, j, k$ are equal to one another. What you've written is ambiguous, and hence meaningless.
– amWhy
Aug 29 at 17:36