Help understand part of the proof. Radius of convergence is $frac{1}{limsup |a_n|^{1/n}}$
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Can you help me understand the highlighted parts of the proof. Thanks :)
Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.
Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.
Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.
For the first highlighted part, I don't understand where the geometric series part comes from.
$sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?
And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?
Thanks!
complex-analysis power-series limsup-and-liminf
asked Oct 18 '14 at 21:07
user181662user181662
674
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add a comment |
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Can you help me understand the highlighted parts of the proof. Thanks :)
Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.
Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.
Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.
For the first highlighted part, I don't understand where the geometric series part comes from.
$sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?
And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?
Thanks!
complex-analysis power-series limsup-and-liminf
asked Oct 18 '14 at 21:07
user181662user181662
674
$endgroup$
add a comment |
$begingroup$
Can you help me understand the highlighted parts of the proof. Thanks :)
Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.
Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.
Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.
For the first highlighted part, I don't understand where the geometric series part comes from.
$sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?
And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?
Thanks!
complex-analysis power-series limsup-and-liminf
asked Oct 18 '14 at 21:07
user181662user181662
674
$endgroup$
Can you help me understand the highlighted parts of the proof. Thanks :)
Theorem: Let $sum{a_nz^n}$ be a power series, let r be its radius of convergence. Then $frac{1}{r} = limsup |a_n|^{1/n}$.
Proof: Let $t = limsup |a_n|^{1/n}$. Suppose first that $t ne 0, infty$. Give $epsilon > 0$, there exits only a finite number of n such that $|a_n|^{1/n} ge t + epsilon$. Thus for all but a finite number of n, we have $|a_n| le (t +epsilon)^n$, whence the series $sum{a_nz^n}$ converges absolutely if $|z| < frac{1}{t+epsilon}$, by comparison with the geometric series. Therefore the radius of convergence $r$ satisfies $r ge frac{1}{t+epsilon}$ for every $epsilon > 0$, whence $rge frac{1}{t}$.
Conversely, given $epsilon$ there exists infinitely many n such that $|a_n|^{1/n} ge t-epsilon$ and therefore, $|a_n|ge (t-epsilon)^n$.
Hence the series $sum{a_nz^n}$ does not converge if $|z| = frac{1}{t-epsilon}$, because it's $n^{th}$ term doesn't even tend to 0. Therefore $r le frac{1}{t-epsilon}$ for every $epsilon > 0$, whence $rle frac{1}{t}$. This concludes the proof for $t ne 0,infty$.
For the first highlighted part, I don't understand where the geometric series part comes from.
$sum{|a_n||z^n|} le sum{(1+epsilon)^nfrac{1}{(t+epsilon)^n}} = sum1$, am I doing this wrong?
And on the second part why wouldn't the equality with $frac{1}{t-epsilon}$ make it converge?
Thanks!
complex-analysis power-series limsup-and-liminf
complex-analysis power-series limsup-and-liminf
asked Oct 18 '14 at 21:07
user181662user181662
674
asked Oct 18 '14 at 21:07
user181662user181662
674
edited Oct 19 '14 at 16:48
user181662
asked Oct 18 '14 at 21:07
user181662user181662
674
asked Oct 18 '14 at 21:07
user181662user181662
674
asked Oct 18 '14 at 21:07
user181662user181662
674
674
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1 Answer
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The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.
To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore
answered Dec 29 '18 at 14:53
RM777RM777
38312
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1 Answer
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1 Answer
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active
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$begingroup$
The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.
To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore
answered Dec 29 '18 at 14:53
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.
To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore
answered Dec 29 '18 at 14:53
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.
To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore
answered Dec 29 '18 at 14:53
RM777RM777
38312
$endgroup$
The Geometrie Series is $sum_{n=1}^{infty}q^n$ If your z is smaller than $frac{1}{t–epsilon}$ Then your q is smaller than 1 which means it Converges and behause it is a Majorat i.e. a bigger series which converges.
To your Second question if a Series converges than the nth term must converge to 0 because otherwise your series would not be cauchy anymore
answered Dec 29 '18 at 14:53
RM777RM777
38312
answered Dec 29 '18 at 14:53
RM777RM777
38312
answered Dec 29 '18 at 14:53
RM777RM777
38312
answered Dec 29 '18 at 14:53
RM777RM777
38312
38312
add a comment |
add a comment |
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