Mixing
Conflictions about Limits
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I just learned the definition of limits, and I learned that if $a_n, b_n $ converges, then $$lim_nto infty (a_n+b_n)=lim_n to infty a_n+lim_n to inftyb_n$$ holds.
And my teacher said that $lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_nto inftyfracfracn(n+1)2n^2=frac12$.
But can't we compute like
$$lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_ntoinftyfrac1n^2+lim_ntoinftyfrac2n^2+lim_ntoinftyfrac3n^2+cdots +lim_ntoinftyfracnn^2=0+0+cdots+0=0$$?
calculus limits
asked 31 mins ago
lminsl
705
add a comment |Â
up vote
3
down vote
favorite
I just learned the definition of limits, and I learned that if $a_n, b_n $ converges, then $$lim_nto infty (a_n+b_n)=lim_n to infty a_n+lim_n to inftyb_n$$ holds.
And my teacher said that $lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_nto inftyfracfracn(n+1)2n^2=frac12$.
But can't we compute like
$$lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_ntoinftyfrac1n^2+lim_ntoinftyfrac2n^2+lim_ntoinftyfrac3n^2+cdots +lim_ntoinftyfracnn^2=0+0+cdots+0=0$$?
calculus limits
asked 31 mins ago
lminsl
705
No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I just learned the definition of limits, and I learned that if $a_n, b_n $ converges, then $$lim_nto infty (a_n+b_n)=lim_n to infty a_n+lim_n to inftyb_n$$ holds.
And my teacher said that $lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_nto inftyfracfracn(n+1)2n^2=frac12$.
But can't we compute like
$$lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_ntoinftyfrac1n^2+lim_ntoinftyfrac2n^2+lim_ntoinftyfrac3n^2+cdots +lim_ntoinftyfracnn^2=0+0+cdots+0=0$$?
calculus limits
asked 31 mins ago
lminsl
705
I just learned the definition of limits, and I learned that if $a_n, b_n $ converges, then $$lim_nto infty (a_n+b_n)=lim_n to infty a_n+lim_n to inftyb_n$$ holds.
And my teacher said that $lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_nto inftyfracfracn(n+1)2n^2=frac12$.
But can't we compute like
$$lim_n to inftyfrac1+2+3+ cdots +nn^2=lim_ntoinftyfrac1n^2+lim_ntoinftyfrac2n^2+lim_ntoinftyfrac3n^2+cdots +lim_ntoinftyfracnn^2=0+0+cdots+0=0$$?
calculus limits
calculus limits
asked 31 mins ago
lminsl
705
asked 31 mins ago
lminsl
705
asked 31 mins ago
lminsl
705
asked 31 mins ago
lminsl
705
asked 31 mins ago
lminsl
705
705
No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago
add a comment |Â
No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago
No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago
No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago
add a comment |Â
3 Answers
3
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up vote
3
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No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that
$$
1 + frac 12 + frac 13 +cdots = +infty
$$
while
$$
1 +frac 12^2+ frac 13^2+ cdots = frac pi^26 in Bbb R.
$$
The theory of series and summation is important in calculus.
UPDATE
Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.
answered 25 mins ago
xbh
4,035320
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
add a comment |Â
up vote
1
down vote
No, because what you learned was that$$lim_ntoinfty(a_n+b_n)=lim_ntoinftya_n+lim_ntoinftyb_n.$$From this, you can deduce that if you have $k$ sequences $bigl(a(i)bigr)_ninmathbb N$, with $iin1,2,ldots,k$, then$$lim_ntoinftybigl(a(1)_n+a(2)_n+cdots+a(k)_nbigr)=lim_ntoinftya(1)_n+lim_ntoinftya(2)_n+cdots+lim_ntoinftya(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.
answered 24 mins ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
down vote
To understand why cannot do that consider a different simpler example: $frac 1 n +frac 1 n+...+frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.
answered 18 mins ago
Kavi Rama Murthy
32k31644
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that
$$
1 + frac 12 + frac 13 +cdots = +infty
$$
while
$$
1 +frac 12^2+ frac 13^2+ cdots = frac pi^26 in Bbb R.
$$
The theory of series and summation is important in calculus.
UPDATE
Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.
answered 25 mins ago
xbh
4,035320
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
add a comment |Â
up vote
3
down vote
No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that
$$
1 + frac 12 + frac 13 +cdots = +infty
$$
while
$$
1 +frac 12^2+ frac 13^2+ cdots = frac pi^26 in Bbb R.
$$
The theory of series and summation is important in calculus.
UPDATE
Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.
answered 25 mins ago
xbh
4,035320
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that
$$
1 + frac 12 + frac 13 +cdots = +infty
$$
while
$$
1 +frac 12^2+ frac 13^2+ cdots = frac pi^26 in Bbb R.
$$
The theory of series and summation is important in calculus.
UPDATE
Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.
answered 25 mins ago
xbh
4,035320
No. The arithmetic law you cited could only allow you to break the limit of sum into sum of limits when there are finitely many summands. For infinite sums, the theory about infinite series would be developed later in your course. You would see that
$$
1 + frac 12 + frac 13 +cdots = +infty
$$
while
$$
1 +frac 12^2+ frac 13^2+ cdots = frac pi^26 in Bbb R.
$$
The theory of series and summation is important in calculus.
UPDATE
Thanks to @MPW. When I say "finitely many summands", I actually mean "a fixed number of summands". I thought the "fixed number" is implied, buy actually my statement does not have such meaning.
answered 25 mins ago
xbh
4,035320
edited 12 mins ago
answered 25 mins ago
xbh
4,035320
answered 25 mins ago
xbh
4,035320
answered 25 mins ago
xbh
4,035320
4,035320
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
add a comment |Â
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
1
1
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
Not quite true. The cited law is for a fixed number of summands. Note that the example has a finite number of summands ($n$ of them); the difficulty is that the number of summands is not bounded as $n$ increases. At no point are you “adding an infinite number of terms.†As I frequently point out, the expression $a_1+a_2+a_3+cdots$ is not a sum — it is a limit.
– MPW
17 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
@MPW Thanks for clarification. I might use the wrong vocabulary.
– xbh
15 mins ago
add a comment |Â
up vote
1
down vote
No, because what you learned was that$$lim_ntoinfty(a_n+b_n)=lim_ntoinftya_n+lim_ntoinftyb_n.$$From this, you can deduce that if you have $k$ sequences $bigl(a(i)bigr)_ninmathbb N$, with $iin1,2,ldots,k$, then$$lim_ntoinftybigl(a(1)_n+a(2)_n+cdots+a(k)_nbigr)=lim_ntoinftya(1)_n+lim_ntoinftya(2)_n+cdots+lim_ntoinftya(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.
answered 24 mins ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
down vote
No, because what you learned was that$$lim_ntoinfty(a_n+b_n)=lim_ntoinftya_n+lim_ntoinftyb_n.$$From this, you can deduce that if you have $k$ sequences $bigl(a(i)bigr)_ninmathbb N$, with $iin1,2,ldots,k$, then$$lim_ntoinftybigl(a(1)_n+a(2)_n+cdots+a(k)_nbigr)=lim_ntoinftya(1)_n+lim_ntoinftya(2)_n+cdots+lim_ntoinftya(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.
answered 24 mins ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, because what you learned was that$$lim_ntoinfty(a_n+b_n)=lim_ntoinftya_n+lim_ntoinftyb_n.$$From this, you can deduce that if you have $k$ sequences $bigl(a(i)bigr)_ninmathbb N$, with $iin1,2,ldots,k$, then$$lim_ntoinftybigl(a(1)_n+a(2)_n+cdots+a(k)_nbigr)=lim_ntoinftya(1)_n+lim_ntoinftya(2)_n+cdots+lim_ntoinftya(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.
answered 24 mins ago
José Carlos Santos
131k17106192
No, because what you learned was that$$lim_ntoinfty(a_n+b_n)=lim_ntoinftya_n+lim_ntoinftyb_n.$$From this, you can deduce that if you have $k$ sequences $bigl(a(i)bigr)_ninmathbb N$, with $iin1,2,ldots,k$, then$$lim_ntoinftybigl(a(1)_n+a(2)_n+cdots+a(k)_nbigr)=lim_ntoinftya(1)_n+lim_ntoinftya(2)_n+cdots+lim_ntoinftya(k)_n.$$But you can't jump from that to infinitely many sequences, which is what you did.
answered 24 mins ago
José Carlos Santos
131k17106192
answered 24 mins ago
José Carlos Santos
131k17106192
answered 24 mins ago
José Carlos Santos
131k17106192
answered 24 mins ago
José Carlos Santos
131k17106192
131k17106192
add a comment |Â
add a comment |Â
up vote
1
down vote
To understand why cannot do that consider a different simpler example: $frac 1 n +frac 1 n+...+frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.
answered 18 mins ago
Kavi Rama Murthy
32k31644
add a comment |Â
up vote
1
down vote
To understand why cannot do that consider a different simpler example: $frac 1 n +frac 1 n+...+frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.
answered 18 mins ago
Kavi Rama Murthy
32k31644
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To understand why cannot do that consider a different simpler example: $frac 1 n +frac 1 n+...+frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.
answered 18 mins ago
Kavi Rama Murthy
32k31644
To understand why cannot do that consider a different simpler example: $frac 1 n +frac 1 n+...+frac 1 n$ ($n$ terms) $=1$. If you take limits the way you did you would get $0+0+cdots +0=1$, which is not true. You can take limits term by term when there are a fixed number of terms but what you have is variable number of terms.
answered 18 mins ago
Kavi Rama Murthy
32k31644
answered 18 mins ago
Kavi Rama Murthy
32k31644
answered 18 mins ago
Kavi Rama Murthy
32k31644
answered 18 mins ago
Kavi Rama Murthy
32k31644
32k31644
add a comment |Â
add a comment |Â
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No you cannot do that. The reason is that you have a non constant number of terms in your sum snd so the limit of the sum is not the sum of the limits.@Iminsl
– AnyAD
26 mins ago