Compute upper limit of $(c_n)^{1/n}$ if $c_{2n}=0$ and $c_{2n+1}=1/(2n+1)$
$begingroup$
I want to find the upper limit of $left((c_n)_{ninmathbb{N}}right)^{frac{1}{n}}$, where $c_n=0$ for $nin 2mathbb{N}$ and $c_n=frac{1}{n}$ for $nin 2mathbb{N}+1$. I know that the upper limit is 1, but how can i prove it?
My idea was to establish $phi:mathbb{N} to mathbb{N}$, $phi(n)=2n+1$ and then I have all my nonzero elements, so that $c_{phi(n)}$=$frac{1}{phi(n)}$
Thanks for your help!
sequences-and-series
edited Dec 13 '18 at 14:15
Did
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
$endgroup$
add a comment |
$begingroup$
I want to find the upper limit of $left((c_n)_{ninmathbb{N}}right)^{frac{1}{n}}$, where $c_n=0$ for $nin 2mathbb{N}$ and $c_n=frac{1}{n}$ for $nin 2mathbb{N}+1$. I know that the upper limit is 1, but how can i prove it?
My idea was to establish $phi:mathbb{N} to mathbb{N}$, $phi(n)=2n+1$ and then I have all my nonzero elements, so that $c_{phi(n)}$=$frac{1}{phi(n)}$
Thanks for your help!
sequences-and-series
edited Dec 13 '18 at 14:15
Did
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
$endgroup$
$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14
add a comment |
$begingroup$
I want to find the upper limit of $left((c_n)_{ninmathbb{N}}right)^{frac{1}{n}}$, where $c_n=0$ for $nin 2mathbb{N}$ and $c_n=frac{1}{n}$ for $nin 2mathbb{N}+1$. I know that the upper limit is 1, but how can i prove it?
My idea was to establish $phi:mathbb{N} to mathbb{N}$, $phi(n)=2n+1$ and then I have all my nonzero elements, so that $c_{phi(n)}$=$frac{1}{phi(n)}$
Thanks for your help!
sequences-and-series
edited Dec 13 '18 at 14:15
Did
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
$endgroup$
I want to find the upper limit of $left((c_n)_{ninmathbb{N}}right)^{frac{1}{n}}$, where $c_n=0$ for $nin 2mathbb{N}$ and $c_n=frac{1}{n}$ for $nin 2mathbb{N}+1$. I know that the upper limit is 1, but how can i prove it?
My idea was to establish $phi:mathbb{N} to mathbb{N}$, $phi(n)=2n+1$ and then I have all my nonzero elements, so that $c_{phi(n)}$=$frac{1}{phi(n)}$
Thanks for your help!
sequences-and-series
sequences-and-series
edited Dec 13 '18 at 14:15
Did
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
edited Dec 13 '18 at 14:15
Did
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
edited Dec 13 '18 at 14:15
Did
248k23224462
edited Dec 13 '18 at 14:15
Did
248k23224462
edited Dec 13 '18 at 14:15
Did
248k23224462
248k23224462
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
asked Dec 13 '18 at 13:24
Michael MaierMichael Maier
859
859
$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14
add a comment |
$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14
$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14
$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
We have that $$(c_n)^frac1n le frac1{sqrt[n]n}le 1$$
is strictly increasing for $nge 2$ and for $n=2k+1$
$$(c_n)^frac1n=frac1{sqrt[2k+1]{2k+1}}to 1$$
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have that $$(c_n)^frac1n le frac1{sqrt[n]n}le 1$$
is strictly increasing for $nge 2$ and for $n=2k+1$
$$(c_n)^frac1n=frac1{sqrt[2k+1]{2k+1}}to 1$$
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that $$(c_n)^frac1n le frac1{sqrt[n]n}le 1$$
is strictly increasing for $nge 2$ and for $n=2k+1$
$$(c_n)^frac1n=frac1{sqrt[2k+1]{2k+1}}to 1$$
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that $$(c_n)^frac1n le frac1{sqrt[n]n}le 1$$
is strictly increasing for $nge 2$ and for $n=2k+1$
$$(c_n)^frac1n=frac1{sqrt[2k+1]{2k+1}}to 1$$
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
$endgroup$
We have that $$(c_n)^frac1n le frac1{sqrt[n]n}le 1$$
is strictly increasing for $nge 2$ and for $n=2k+1$
$$(c_n)^frac1n=frac1{sqrt[2k+1]{2k+1}}to 1$$
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
answered Dec 13 '18 at 13:31
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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$begingroup$
Your notations are off. You are asking for the "upper limit" (that is, the limsup) of $(c_n)^{1/n}$.
$endgroup$
– Did
Dec 13 '18 at 14:14