Mixing
Show that if $sum_{k=0}^{infty}a_kz^k$ converges for $zinmathbb{C}$ then $sum_{k=0}^{infty}a_{k+n}z^k$ also...
$begingroup$
Attention: $|z|<R$ where $R$ is the radius of convergence
Show that if $sum_{k=0}^{infty}a_kz^k$ converges for $zinmathbb{C}$ then $sum_{k=0}^{infty}a_{k+n}z^k$ also converges $ninmathbb{N}$
I have tried to solve this Problem for three Hours now here are my results so far but Nothing seems to work out:
I know since $|z|<R$ that the series converges absolutely.
A necessary condition for the convergence of a series is that the secuence of ist summands must be convergin to 0.
Therefore:
$lim_{krightarrow infty} |a_kz^k|=0$
Because the elements of the sequence are all positive the sequence must decrease at some Point.
That is without loss of generality that
$$|a_kz^k|geq|a_{k+1}z^{k+1}|geq|a_{k+2}z^{k+2}| geq …geq |a_{k+n}z^{k+n}|$$
Implying $frac{|a_{k+n}z^{k+n}|}{|a_kz^k|}leq 1iff frac{a_{k+n}}{a_k}leq frac{1}{z^n}$
Therefore
$sum_{k=0}^{infty}a_{k+n}z^k= sum_{k=0}^{infty} a_kfrac{a_k+n}{a_k}z^kleq sum_{k=0}^{infty} |a_kfrac{a_k+n}{a_k}z^k|leq sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|$
And here is the Problem If I now say that
$sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|leq sum_{k=0}^{infty}|a_kz^k|$
Then I am assuming that $|z^{k-n}|leq |z^k|iff 1<|z^n|$
Therefore this does not work generally.
Please help me.
complex-analysis power-series
asked Feb 3 at 13:04
New2MathNew2Math
31015
$endgroup$
add a comment |
$begingroup$
Attention: $|z|<R$ where $R$ is the radius of convergence
Show that if $sum_{k=0}^{infty}a_kz^k$ converges for $zinmathbb{C}$ then $sum_{k=0}^{infty}a_{k+n}z^k$ also converges $ninmathbb{N}$
I have tried to solve this Problem for three Hours now here are my results so far but Nothing seems to work out:
I know since $|z|<R$ that the series converges absolutely.
A necessary condition for the convergence of a series is that the secuence of ist summands must be convergin to 0.
Therefore:
$lim_{krightarrow infty} |a_kz^k|=0$
Because the elements of the sequence are all positive the sequence must decrease at some Point.
That is without loss of generality that
$$|a_kz^k|geq|a_{k+1}z^{k+1}|geq|a_{k+2}z^{k+2}| geq …geq |a_{k+n}z^{k+n}|$$
Implying $frac{|a_{k+n}z^{k+n}|}{|a_kz^k|}leq 1iff frac{a_{k+n}}{a_k}leq frac{1}{z^n}$
Therefore
$sum_{k=0}^{infty}a_{k+n}z^k= sum_{k=0}^{infty} a_kfrac{a_k+n}{a_k}z^kleq sum_{k=0}^{infty} |a_kfrac{a_k+n}{a_k}z^k|leq sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|$
And here is the Problem If I now say that
$sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|leq sum_{k=0}^{infty}|a_kz^k|$
Then I am assuming that $|z^{k-n}|leq |z^k|iff 1<|z^n|$
Therefore this does not work generally.
Please help me.
complex-analysis power-series
asked Feb 3 at 13:04
New2MathNew2Math
31015
$endgroup$
$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19
add a comment |
$begingroup$
Attention: $|z|<R$ where $R$ is the radius of convergence
Show that if $sum_{k=0}^{infty}a_kz^k$ converges for $zinmathbb{C}$ then $sum_{k=0}^{infty}a_{k+n}z^k$ also converges $ninmathbb{N}$
I have tried to solve this Problem for three Hours now here are my results so far but Nothing seems to work out:
I know since $|z|<R$ that the series converges absolutely.
A necessary condition for the convergence of a series is that the secuence of ist summands must be convergin to 0.
Therefore:
$lim_{krightarrow infty} |a_kz^k|=0$
Because the elements of the sequence are all positive the sequence must decrease at some Point.
That is without loss of generality that
$$|a_kz^k|geq|a_{k+1}z^{k+1}|geq|a_{k+2}z^{k+2}| geq …geq |a_{k+n}z^{k+n}|$$
Implying $frac{|a_{k+n}z^{k+n}|}{|a_kz^k|}leq 1iff frac{a_{k+n}}{a_k}leq frac{1}{z^n}$
Therefore
$sum_{k=0}^{infty}a_{k+n}z^k= sum_{k=0}^{infty} a_kfrac{a_k+n}{a_k}z^kleq sum_{k=0}^{infty} |a_kfrac{a_k+n}{a_k}z^k|leq sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|$
And here is the Problem If I now say that
$sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|leq sum_{k=0}^{infty}|a_kz^k|$
Then I am assuming that $|z^{k-n}|leq |z^k|iff 1<|z^n|$
Therefore this does not work generally.
Please help me.
complex-analysis power-series
asked Feb 3 at 13:04
New2MathNew2Math
31015
$endgroup$
Attention: $|z|<R$ where $R$ is the radius of convergence
Show that if $sum_{k=0}^{infty}a_kz^k$ converges for $zinmathbb{C}$ then $sum_{k=0}^{infty}a_{k+n}z^k$ also converges $ninmathbb{N}$
I have tried to solve this Problem for three Hours now here are my results so far but Nothing seems to work out:
I know since $|z|<R$ that the series converges absolutely.
A necessary condition for the convergence of a series is that the secuence of ist summands must be convergin to 0.
Therefore:
$lim_{krightarrow infty} |a_kz^k|=0$
Because the elements of the sequence are all positive the sequence must decrease at some Point.
That is without loss of generality that
$$|a_kz^k|geq|a_{k+1}z^{k+1}|geq|a_{k+2}z^{k+2}| geq …geq |a_{k+n}z^{k+n}|$$
Implying $frac{|a_{k+n}z^{k+n}|}{|a_kz^k|}leq 1iff frac{a_{k+n}}{a_k}leq frac{1}{z^n}$
Therefore
$sum_{k=0}^{infty}a_{k+n}z^k= sum_{k=0}^{infty} a_kfrac{a_k+n}{a_k}z^kleq sum_{k=0}^{infty} |a_kfrac{a_k+n}{a_k}z^k|leq sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|$
And here is the Problem If I now say that
$sum_{k=0}^{infty}|a_kfrac{1}{z^n}z^k|leq sum_{k=0}^{infty}|a_kz^k|$
Then I am assuming that $|z^{k-n}|leq |z^k|iff 1<|z^n|$
Therefore this does not work generally.
Please help me.
complex-analysis power-series
complex-analysis power-series
asked Feb 3 at 13:04
New2MathNew2Math
31015
asked Feb 3 at 13:04
New2MathNew2Math
31015
asked Feb 3 at 13:04
New2MathNew2Math
31015
asked Feb 3 at 13:04
New2MathNew2Math
31015
asked Feb 3 at 13:04
New2MathNew2Math
31015
31015
$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19
add a comment |
$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19
$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19
$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19
add a comment |
1 Answer
1
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oldest
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$begingroup$
If $sum_{k=0}^{infty}a_kz^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$. Thus also $sum_{k=0}^{infty}a_{k+n}z^k$ for $znot=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $sum_{k=0}^{infty}a_{k+n}z^k$ is at least that of $sum_{k=0}^{infty}a_kz^k$. If, on
the other hand, $sum_{k=0}^{infty}a_{k+n}z^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$ and therefore also $sum_{k=0}^{infty}a_kz^k=sum_{k=0}^{n-1
}a_kz^k+sum_{k=0}^{infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $sum_{k=0}^{infty}a_kz^k$ is at least that of $sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
Thus they are equal.
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
$endgroup$
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
add a comment |
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$begingroup$
If $sum_{k=0}^{infty}a_kz^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$. Thus also $sum_{k=0}^{infty}a_{k+n}z^k$ for $znot=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $sum_{k=0}^{infty}a_{k+n}z^k$ is at least that of $sum_{k=0}^{infty}a_kz^k$. If, on
the other hand, $sum_{k=0}^{infty}a_{k+n}z^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$ and therefore also $sum_{k=0}^{infty}a_kz^k=sum_{k=0}^{n-1
}a_kz^k+sum_{k=0}^{infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $sum_{k=0}^{infty}a_kz^k$ is at least that of $sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
Thus they are equal.
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
$endgroup$
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
add a comment |
$begingroup$
If $sum_{k=0}^{infty}a_kz^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$. Thus also $sum_{k=0}^{infty}a_{k+n}z^k$ for $znot=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $sum_{k=0}^{infty}a_{k+n}z^k$ is at least that of $sum_{k=0}^{infty}a_kz^k$. If, on
the other hand, $sum_{k=0}^{infty}a_{k+n}z^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$ and therefore also $sum_{k=0}^{infty}a_kz^k=sum_{k=0}^{n-1
}a_kz^k+sum_{k=0}^{infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $sum_{k=0}^{infty}a_kz^k$ is at least that of $sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
Thus they are equal.
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
$endgroup$
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
add a comment |
$begingroup$
If $sum_{k=0}^{infty}a_kz^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$. Thus also $sum_{k=0}^{infty}a_{k+n}z^k$ for $znot=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $sum_{k=0}^{infty}a_{k+n}z^k$ is at least that of $sum_{k=0}^{infty}a_kz^k$. If, on
the other hand, $sum_{k=0}^{infty}a_{k+n}z^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$ and therefore also $sum_{k=0}^{infty}a_kz^k=sum_{k=0}^{n-1
}a_kz^k+sum_{k=0}^{infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $sum_{k=0}^{infty}a_kz^k$ is at least that of $sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
Thus they are equal.
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
$endgroup$
If $sum_{k=0}^{infty}a_kz^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$. Thus also $sum_{k=0}^{infty}a_{k+n}z^k$ for $znot=0$, but trivially also for $z=0$. Thus the radius of converges for the power series $sum_{k=0}^{infty}a_{k+n}z^k$ is at least that of $sum_{k=0}^{infty}a_kz^k$. If, on
the other hand, $sum_{k=0}^{infty}a_{k+n}z^k$ converges absolutely so does $sum_{k=0}^{infty}a_{k+n}z^{k+n}$ and therefore also $sum_{k=0}^{infty}a_kz^k=sum_{k=0}^{n-1
}a_kz^k+sum_{k=0}^{infty}a_{k+n}z^{k+n}$. This implies that the radius of convergence of $sum_{k=0}^{infty}a_kz^k$ is at least that of $sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
Thus they are equal.
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
answered Feb 3 at 13:34
Jens SchwaigerJens Schwaiger
1,639138
1,639138
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
add a comment |
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
I don't understand why $z neq 0$ Because $sum_{k=0}^{infty}a_{k+n}z^{k+n}=sum_{k=0}^{infty}a_{k+n}z^kz^n=z^nsum_{k=0}^{infty}a_{k+n}z^k$, why is it not allowed that $z$ is $0$?
$endgroup$
– New2Math
Feb 3 at 14:01
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
$begingroup$
The other direction is important. There we divide by $z^n$: $sum_{k=0}^{infty}a_{k+n}z^k=frac1{z^n} sum_{k=0}^{infty}a_{k+n}z^{k+n}$.
$endgroup$
– Jens Schwaiger
Feb 3 at 22:08
add a comment |
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$begingroup$
A related question was asked recently: math.stackexchange.com/questions/3098468/….
$endgroup$
– Martin R
Feb 3 at 13:19