What is the Convergence Radius and Domain of $sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$?
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So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$
I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.
It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .
Thanks.
calculus sequences-and-series convergence
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|
show 3 more comments
$begingroup$
So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$
I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.
It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .
Thanks.
calculus sequences-and-series convergence
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4
$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
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– Wojowu
Dec 27 '18 at 17:04
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About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
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– Jakobian
Dec 27 '18 at 17:13
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@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56
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@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
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– Roee
Dec 27 '18 at 17:57
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@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58
|
show 3 more comments
$begingroup$
So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$
I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.
It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .
Thanks.
calculus sequences-and-series convergence
$endgroup$
So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$
I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.
It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .
Thanks.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Dec 27 '18 at 19:28
Roee
asked Dec 27 '18 at 17:00
RoeeRoee
374
374
4
$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04
$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13
$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56
$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57
$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58
|
show 3 more comments
4
$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04
$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13
$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56
$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57
$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58
4
4
$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04
$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04
$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13
$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13
$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56
$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56
$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57
$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57
$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58
$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.
The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.
$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.
Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
$endgroup$
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
add a comment |
$begingroup$
Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.
$endgroup$
add a comment |
$begingroup$
i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.
The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.
$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.
Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
$endgroup$
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
add a comment |
$begingroup$
For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.
The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.
$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.
Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
$endgroup$
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
add a comment |
$begingroup$
For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.
The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.
$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.
Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
$endgroup$
For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.
The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.
$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$
The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.
For any point $z$ on the unit circle, we have
$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$
The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.
Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.
This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).
edited Dec 27 '18 at 19:10
answered Dec 27 '18 at 18:47
achille huiachille hui
96.3k5132261
96.3k5132261
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
add a comment |
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
2
2
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
$begingroup$
A typo, Fabry gap theorem instead of Fabry gay theorem
$endgroup$
– Jakobian
Dec 27 '18 at 19:09
1
1
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
$begingroup$
@Jakobian oops, what a mistake.
$endgroup$
– achille hui
Dec 27 '18 at 19:11
add a comment |
$begingroup$
Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.
$endgroup$
add a comment |
$begingroup$
Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.
$endgroup$
add a comment |
$begingroup$
Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.
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Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.
answered Dec 27 '18 at 19:48
ZacharyZachary
2,3751214
2,3751214
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i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
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i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
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add a comment |
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i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
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i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$
answered Dec 27 '18 at 20:08
zhw.zhw.
74.3k43175
74.3k43175
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Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
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– Wojowu
Dec 27 '18 at 17:04
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About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
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– Jakobian
Dec 27 '18 at 17:13
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@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
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– Roee
Dec 27 '18 at 17:56
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@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
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– Roee
Dec 27 '18 at 17:57
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@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
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– Jakobian
Dec 27 '18 at 17:58