Prove the shift rule for series
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Let $N$ be a natural number. Then the series $sum_{n=1}^{infty}a_n$ converges if and only if the series $sum_{n=1}^{infty}a_{N+n}$ converges.
I've tried splitting $sum_{n=1}^{infty}a_n$ into $sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$ but am not getting anywhere. I've also tried using the $epsilon$ definition for convergence but am getting confused.
sequences-and-series proof-writing
asked Nov 17 at 18:23
m.bazza
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add a comment |
up vote
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favorite
Let $N$ be a natural number. Then the series $sum_{n=1}^{infty}a_n$ converges if and only if the series $sum_{n=1}^{infty}a_{N+n}$ converges.
I've tried splitting $sum_{n=1}^{infty}a_n$ into $sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$ but am not getting anywhere. I've also tried using the $epsilon$ definition for convergence but am getting confused.
sequences-and-series proof-writing
asked Nov 17 at 18:23
m.bazza
827
2
Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
How would I go about this?
– m.bazza
Nov 17 at 18:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $N$ be a natural number. Then the series $sum_{n=1}^{infty}a_n$ converges if and only if the series $sum_{n=1}^{infty}a_{N+n}$ converges.
I've tried splitting $sum_{n=1}^{infty}a_n$ into $sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$ but am not getting anywhere. I've also tried using the $epsilon$ definition for convergence but am getting confused.
sequences-and-series proof-writing
asked Nov 17 at 18:23
m.bazza
827
Let $N$ be a natural number. Then the series $sum_{n=1}^{infty}a_n$ converges if and only if the series $sum_{n=1}^{infty}a_{N+n}$ converges.
I've tried splitting $sum_{n=1}^{infty}a_n$ into $sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$ but am not getting anywhere. I've also tried using the $epsilon$ definition for convergence but am getting confused.
sequences-and-series proof-writing
sequences-and-series proof-writing
asked Nov 17 at 18:23
m.bazza
827
asked Nov 17 at 18:23
m.bazza
827
asked Nov 17 at 18:23
m.bazza
827
asked Nov 17 at 18:23
m.bazza
827
asked Nov 17 at 18:23
m.bazza
827
827
2
Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
How would I go about this?
– m.bazza
Nov 17 at 18:42
add a comment |
2
Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
How would I go about this?
– m.bazza
Nov 17 at 18:42
2
2
Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
How would I go about this?
– m.bazza
Nov 17 at 18:42
How would I go about this?
– m.bazza
Nov 17 at 18:42
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
$sum_{n=1}^Na_n$ is just a number, let's call it $C$. So if the sum of $a_n$ converges to $A$, then the $$A=sum_{n=1}^infty a_n=sum_{n=1}^N a_n+sum_{n=1}^infty a_{n+N}=C+sum_{n=1}^infty a_{n+N}$$ wich yields $$sum_{n=1}^infty a_{n+N}=A-C$$
Since both $A$ and $C$ are finite then $A-C$ is finite, so the series converges.
Similarly, you can go the other way around
answered Nov 17 at 19:14
Andrei
10.3k21025
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
add a comment |
up vote
0
down vote
As you have written we have $$sum_{n=1}^{infty}a_n=sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$$where $$sum_{n=1}^{N}a_n$$ is bounded therefore $sum_{n=1}^{infty}a_n$ is bounded if and only if $sum_{n=1}^{infty}a_{n+N}$ is bounded.
answered 2 days ago
Mostafa Ayaz
13.1k3735
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$sum_{n=1}^Na_n$ is just a number, let's call it $C$. So if the sum of $a_n$ converges to $A$, then the $$A=sum_{n=1}^infty a_n=sum_{n=1}^N a_n+sum_{n=1}^infty a_{n+N}=C+sum_{n=1}^infty a_{n+N}$$ wich yields $$sum_{n=1}^infty a_{n+N}=A-C$$
Since both $A$ and $C$ are finite then $A-C$ is finite, so the series converges.
Similarly, you can go the other way around
answered Nov 17 at 19:14
Andrei
10.3k21025
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
add a comment |
up vote
1
down vote
$sum_{n=1}^Na_n$ is just a number, let's call it $C$. So if the sum of $a_n$ converges to $A$, then the $$A=sum_{n=1}^infty a_n=sum_{n=1}^N a_n+sum_{n=1}^infty a_{n+N}=C+sum_{n=1}^infty a_{n+N}$$ wich yields $$sum_{n=1}^infty a_{n+N}=A-C$$
Since both $A$ and $C$ are finite then $A-C$ is finite, so the series converges.
Similarly, you can go the other way around
answered Nov 17 at 19:14
Andrei
10.3k21025
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
add a comment |
up vote
1
down vote
up vote
1
down vote
$sum_{n=1}^Na_n$ is just a number, let's call it $C$. So if the sum of $a_n$ converges to $A$, then the $$A=sum_{n=1}^infty a_n=sum_{n=1}^N a_n+sum_{n=1}^infty a_{n+N}=C+sum_{n=1}^infty a_{n+N}$$ wich yields $$sum_{n=1}^infty a_{n+N}=A-C$$
Since both $A$ and $C$ are finite then $A-C$ is finite, so the series converges.
Similarly, you can go the other way around
answered Nov 17 at 19:14
Andrei
10.3k21025
$sum_{n=1}^Na_n$ is just a number, let's call it $C$. So if the sum of $a_n$ converges to $A$, then the $$A=sum_{n=1}^infty a_n=sum_{n=1}^N a_n+sum_{n=1}^infty a_{n+N}=C+sum_{n=1}^infty a_{n+N}$$ wich yields $$sum_{n=1}^infty a_{n+N}=A-C$$
Since both $A$ and $C$ are finite then $A-C$ is finite, so the series converges.
Similarly, you can go the other way around
answered Nov 17 at 19:14
Andrei
10.3k21025
answered Nov 17 at 19:14
Andrei
10.3k21025
answered Nov 17 at 19:14
Andrei
10.3k21025
answered Nov 17 at 19:14
Andrei
10.3k21025
10.3k21025
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
add a comment |
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
ok, but how do you know that $C$ is finite?
– m.bazza
Nov 17 at 19:17
1
1
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
It is a finite sum of given finite numbers.
– Andrei
Nov 17 at 19:31
add a comment |
up vote
0
down vote
As you have written we have $$sum_{n=1}^{infty}a_n=sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$$where $$sum_{n=1}^{N}a_n$$ is bounded therefore $sum_{n=1}^{infty}a_n$ is bounded if and only if $sum_{n=1}^{infty}a_{n+N}$ is bounded.
answered 2 days ago
Mostafa Ayaz
13.1k3735
add a comment |
up vote
0
down vote
As you have written we have $$sum_{n=1}^{infty}a_n=sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$$where $$sum_{n=1}^{N}a_n$$ is bounded therefore $sum_{n=1}^{infty}a_n$ is bounded if and only if $sum_{n=1}^{infty}a_{n+N}$ is bounded.
answered 2 days ago
Mostafa Ayaz
13.1k3735
add a comment |
up vote
0
down vote
up vote
0
down vote
As you have written we have $$sum_{n=1}^{infty}a_n=sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$$where $$sum_{n=1}^{N}a_n$$ is bounded therefore $sum_{n=1}^{infty}a_n$ is bounded if and only if $sum_{n=1}^{infty}a_{n+N}$ is bounded.
answered 2 days ago
Mostafa Ayaz
13.1k3735
As you have written we have $$sum_{n=1}^{infty}a_n=sum_{n=1}^{infty}a_{N+n} +sum_{n=1}^{N}a_n$$where $$sum_{n=1}^{N}a_n$$ is bounded therefore $sum_{n=1}^{infty}a_n$ is bounded if and only if $sum_{n=1}^{infty}a_{n+N}$ is bounded.
answered 2 days ago
Mostafa Ayaz
13.1k3735
answered 2 days ago
Mostafa Ayaz
13.1k3735
answered 2 days ago
Mostafa Ayaz
13.1k3735
answered 2 days ago
Mostafa Ayaz
13.1k3735
13.1k3735
add a comment |
add a comment |
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Think about partial sums.
– Lord Shark the Unknown
Nov 17 at 18:23
How would I go about this?
– m.bazza
Nov 17 at 18:42