Given $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, prove the following properties.
Given $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, prove:
1) If $alpha > 0$, show $sum_{n = 0}^{infty} a_{n}x^{n}$ converges
if $|x| < 1/alpha$ and diverges if $|x| > 1/alpha$
2) If $alpha = 0$, show that $sum_{n = 0}^{infty} a_{n} x^{n}$
converges for all $x neq 0$.
My try:
1) Suppose $|x| < 1/alpha$. Then, since $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = lim_{ntoinfty} |a_{n}|^{1/n}|x| = |x|lim_{ntoinfty}|a_{n}|^{1/n} < frac{1}{alpha} cdot alpha = 1.$$
I don't know how to show convergence from this, though. I also can't show that it diverges if it's greater than $1/alpha$.
2) For $alpha = 0$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = |x|lim_{ntoinfty}|a_{n}|^{1/n} = |x| cdot 0 = 0.$$
Again, I don't know if this is in the right direction.
Any help is appreciated.
real-analysis sequences-and-series limits convergence power-series
asked Dec 10 '18 at 5:21
joseph
4329
add a comment |
Given $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, prove:
1) If $alpha > 0$, show $sum_{n = 0}^{infty} a_{n}x^{n}$ converges
if $|x| < 1/alpha$ and diverges if $|x| > 1/alpha$
2) If $alpha = 0$, show that $sum_{n = 0}^{infty} a_{n} x^{n}$
converges for all $x neq 0$.
My try:
1) Suppose $|x| < 1/alpha$. Then, since $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = lim_{ntoinfty} |a_{n}|^{1/n}|x| = |x|lim_{ntoinfty}|a_{n}|^{1/n} < frac{1}{alpha} cdot alpha = 1.$$
I don't know how to show convergence from this, though. I also can't show that it diverges if it's greater than $1/alpha$.
2) For $alpha = 0$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = |x|lim_{ntoinfty}|a_{n}|^{1/n} = |x| cdot 0 = 0.$$
Again, I don't know if this is in the right direction.
Any help is appreciated.
real-analysis sequences-and-series limits convergence power-series
asked Dec 10 '18 at 5:21
joseph
4329
add a comment |
Given $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, prove:
1) If $alpha > 0$, show $sum_{n = 0}^{infty} a_{n}x^{n}$ converges
if $|x| < 1/alpha$ and diverges if $|x| > 1/alpha$
2) If $alpha = 0$, show that $sum_{n = 0}^{infty} a_{n} x^{n}$
converges for all $x neq 0$.
My try:
1) Suppose $|x| < 1/alpha$. Then, since $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = lim_{ntoinfty} |a_{n}|^{1/n}|x| = |x|lim_{ntoinfty}|a_{n}|^{1/n} < frac{1}{alpha} cdot alpha = 1.$$
I don't know how to show convergence from this, though. I also can't show that it diverges if it's greater than $1/alpha$.
2) For $alpha = 0$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = |x|lim_{ntoinfty}|a_{n}|^{1/n} = |x| cdot 0 = 0.$$
Again, I don't know if this is in the right direction.
Any help is appreciated.
real-analysis sequences-and-series limits convergence power-series
asked Dec 10 '18 at 5:21
joseph
4329
Given $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, prove:
1) If $alpha > 0$, show $sum_{n = 0}^{infty} a_{n}x^{n}$ converges
if $|x| < 1/alpha$ and diverges if $|x| > 1/alpha$
2) If $alpha = 0$, show that $sum_{n = 0}^{infty} a_{n} x^{n}$
converges for all $x neq 0$.
My try:
1) Suppose $|x| < 1/alpha$. Then, since $lim_{ntoinfty} |a_{n}|^{1/n} = alpha$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = lim_{ntoinfty} |a_{n}|^{1/n}|x| = |x|lim_{ntoinfty}|a_{n}|^{1/n} < frac{1}{alpha} cdot alpha = 1.$$
I don't know how to show convergence from this, though. I also can't show that it diverges if it's greater than $1/alpha$.
2) For $alpha = 0$, we have
$$lim_{ntoinfty} |a_{n}x^{n}|^{1/n} = |x|lim_{ntoinfty}|a_{n}|^{1/n} = |x| cdot 0 = 0.$$
Again, I don't know if this is in the right direction.
Any help is appreciated.
real-analysis sequences-and-series limits convergence power-series
real-analysis sequences-and-series limits convergence power-series
asked Dec 10 '18 at 5:21
joseph
4329
asked Dec 10 '18 at 5:21
joseph
4329
asked Dec 10 '18 at 5:21
joseph
4329
asked Dec 10 '18 at 5:21
joseph
4329
asked Dec 10 '18 at 5:21
joseph
4329
4329
add a comment |
add a comment |
1 Answer
1
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oldest
votes
1) is an immediate application of root test.
For 2) let $0 <epsilon <frac 1 {|x|}$. Then $|a_n|^{1/n} <epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|epsilon x|^{n}$. Since $|epsilon x|<1$ the geometric series $sum |epsilon x|^{n}$ converges. Hence the given series also converges.
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
add a comment |
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1 Answer
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1 Answer
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oldest
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1) is an immediate application of root test.
For 2) let $0 <epsilon <frac 1 {|x|}$. Then $|a_n|^{1/n} <epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|epsilon x|^{n}$. Since $|epsilon x|<1$ the geometric series $sum |epsilon x|^{n}$ converges. Hence the given series also converges.
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
add a comment |
1) is an immediate application of root test.
For 2) let $0 <epsilon <frac 1 {|x|}$. Then $|a_n|^{1/n} <epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|epsilon x|^{n}$. Since $|epsilon x|<1$ the geometric series $sum |epsilon x|^{n}$ converges. Hence the given series also converges.
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
add a comment |
1) is an immediate application of root test.
For 2) let $0 <epsilon <frac 1 {|x|}$. Then $|a_n|^{1/n} <epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|epsilon x|^{n}$. Since $|epsilon x|<1$ the geometric series $sum |epsilon x|^{n}$ converges. Hence the given series also converges.
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
1) is an immediate application of root test.
For 2) let $0 <epsilon <frac 1 {|x|}$. Then $|a_n|^{1/n} <epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|epsilon x|^{n}$. Since $|epsilon x|<1$ the geometric series $sum |epsilon x|^{n}$ converges. Hence the given series also converges.
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 5:38
Kavi Rama Murthy
50.4k31854
50.4k31854
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
add a comment |
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
How is $(1)$ an immediate application of the Root Test?
– joseph
Dec 10 '18 at 5:47
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
See en.wikipedia.org/wiki/Root_test
– Kavi Rama Murthy
Dec 10 '18 at 5:48
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
why is it true that $|a_{n}|^{1/n} < epsilon$ for sufficiently large $n$?
– joseph
Dec 10 '18 at 6:50
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
@joseph Because $alpha =0$ in 2).
– Kavi Rama Murthy
Dec 10 '18 at 7:15
add a comment |
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