Easy question on infinite series.
up vote
-1
down vote
favorite
$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
On the one hand partial sums of this series are equal to 0, so our infinite series must converge to 0.
On the other hand
$$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}=sum_{k=0}^{infty} 1^{2k+1}$$
Obviously this series is not converge to 0. And i cant understand what rules i'm breaking?
real-analysis
edited yesterday
Matti P.
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
-1
down vote
favorite
$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
On the one hand partial sums of this series are equal to 0, so our infinite series must converge to 0.
On the other hand
$$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}=sum_{k=0}^{infty} 1^{2k+1}$$
Obviously this series is not converge to 0. And i cant understand what rules i'm breaking?
real-analysis
edited yesterday
Matti P.
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov 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 |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
On the one hand partial sums of this series are equal to 0, so our infinite series must converge to 0.
On the other hand
$$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}=sum_{k=0}^{infty} 1^{2k+1}$$
Obviously this series is not converge to 0. And i cant understand what rules i'm breaking?
real-analysis
edited yesterday
Matti P.
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
On the one hand partial sums of this series are equal to 0, so our infinite series must converge to 0.
On the other hand
$$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}=sum_{k=0}^{infty} 1^{2k+1}$$
Obviously this series is not converge to 0. And i cant understand what rules i'm breaking?
real-analysis
real-analysis
edited yesterday
Matti P.
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Matti P.
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Matti P.
1,625413
edited yesterday
Matti P.
1,625413
edited yesterday
Matti P.
1,625413
1,625413
asked yesterday
A.Kazakov
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
A.Kazakov
32
asked yesterday
A.Kazakov
32
32
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
A.Kazakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A.Kazakov 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 |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}neqsum_{k=0}^{infty} 1^{2k+1}$
I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?
answered yesterday
KnowsNothing
355
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}neqsum_{k=0}^{infty} 1^{2k+1}$
I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?
answered yesterday
KnowsNothing
355
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
add a comment |
up vote
0
down vote
accepted
$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}neqsum_{k=0}^{infty} 1^{2k+1}$
I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?
answered yesterday
KnowsNothing
355
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}neqsum_{k=0}^{infty} 1^{2k+1}$
I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?
answered yesterday
KnowsNothing
355
$sum_{k=0}^{infty} 1^k-1^{2k}= sum_{k=0}^{infty} 1^k-sum_{k=0}^{infty} 1^{2k}neqsum_{k=0}^{infty} 1^{2k+1}$
I do not understand why you think the last expression is equal to your sum. Are you able to explain why you believe that is the case?
answered yesterday
KnowsNothing
355
answered yesterday
KnowsNothing
355
answered yesterday
KnowsNothing
355
answered yesterday
KnowsNothing
355
355
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
add a comment |
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Because the first addend is sum of all degrees of 1 and the second addend is the sum of all even degrees of 1, so their difference is the sum of all odd degrees of 1.
– A.Kazakov
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
Oh I see now. I think this is one of those cases where you can't change the order of your summation. So your original sum was obviously 0 (as you stated), I'm not sure that is correct to split it into two sums. I'll try and find an example of when changing the order of summation produces rubbish
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
I suppose that Yves Daoust's answer to this question math.stackexchange.com/questions/657241/… is a good example of this kind of stuff.
– KnowsNothing
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
Thank you very much. I got it.
– A.Kazakov
yesterday
add a comment |
A.Kazakov is a new contributor. Be nice, and check out our Code of Conduct.
A.Kazakov is a new contributor. Be nice, and check out our Code of Conduct.
A.Kazakov is a new contributor. Be nice, and check out our Code of Conduct.
A.Kazakov is a new contributor. Be nice, and check out our Code of Conduct.
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