Discuss the convergence of the series $sum_{n=2}^infty(ln{n})^{-ln(ln{n})}$
$begingroup$
Discuss the convergence or divergence of the series where the general term $x_n$ is given by
$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$
calculus real-analysis sequences-and-series limits convergence
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
$endgroup$
add a comment |
$begingroup$
Discuss the convergence or divergence of the series where the general term $x_n$ is given by
$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$
calculus real-analysis sequences-and-series limits convergence
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
$endgroup$
$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
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I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
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@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
2
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32
add a comment |
$begingroup$
Discuss the convergence or divergence of the series where the general term $x_n$ is given by
$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$
calculus real-analysis sequences-and-series limits convergence
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
$endgroup$
Discuss the convergence or divergence of the series where the general term $x_n$ is given by
$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$
calculus real-analysis sequences-and-series limits convergence
calculus real-analysis sequences-and-series limits convergence
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
edited Aug 23 '14 at 14:55
Yiorgos S. Smyrlis
62.9k1384163
62.9k1384163
asked Mar 16 '14 at 13:15
anonymousanonymous
211
asked Mar 16 '14 at 13:15
anonymousanonymous
211
asked Mar 16 '14 at 13:15
anonymousanonymous
211
211
$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
2
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32
add a comment |
$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
2
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32
$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
2
2
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
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
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active
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$begingroup$
We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
add a comment |
$begingroup$
We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
add a comment |
$begingroup$
We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
$endgroup$
We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
answered Mar 16 '14 at 13:29
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.9k1384163
62.9k1384163
add a comment |
add a comment |
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$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20
$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26
$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27
$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29
2
$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32