Easy question on infinite series.
up vote
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$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
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 Nov 21 at 8:08
Matti P.
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
add a comment |
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 Nov 21 at 8:08
Matti P.
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
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 Nov 21 at 8:08
Matti P.
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
$$sum_{k=0}^{infty} 1^k-1^{2k}.$$
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 Nov 21 at 8:08
Matti P.
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
edited Nov 21 at 8:08
Matti P.
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
edited Nov 21 at 8:08
Matti P.
1,681413
edited Nov 21 at 8:08
Matti P.
1,681413
edited Nov 21 at 8:08
Matti P.
1,681413
1,681413
asked Nov 21 at 8:03
A.Kazakov
63
asked Nov 21 at 8:03
A.Kazakov
63
asked Nov 21 at 8:03
A.Kazakov
63
63
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 Nov 21 at 8:16
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
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 Nov 21 at 8:16
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
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 Nov 21 at 8:16
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
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 Nov 21 at 8:16
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 Nov 21 at 8:16
KnowsNothing
355
answered Nov 21 at 8:16
KnowsNothing
355
answered Nov 21 at 8:16
KnowsNothing
355
answered Nov 21 at 8:16
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
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
Nov 21 at 8:30
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
Nov 21 at 8:30
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
Nov 21 at 8:37
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
Nov 21 at 8:37
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
Nov 21 at 8:41
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
Nov 21 at 8:41
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
Thank you very much. I got it.
– A.Kazakov
Nov 21 at 18:57
add a comment |
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