For which values of $p$ does $sum_{n=1}^{infty}(-1)^n{{ln^pn}over{n}}$ converge?
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I want to find for which values of $p$ the following series converges: $$sum_{n=1}^{infty}(-1)^n{{ln^pn}over{n}}$$
I found that for any value of $p$, $$lim_{nto infty} {ln^p nover n}=0$$
Now I have to prove that $f:x longmapsto{ln^p xover x}$ is a decreasing function to apply the alternating series test.
Its derivative is $$f'(x)={{pover x}ln^{p-1}x times x-ln^pxover x^2}={ln^{p-1}x(p-ln x)over x^2}$$ Thus, for $x>1,$ $$f'(x)leq0Leftrightarrow pleq ln{x}$$
And then I'm not quite sure whether it means something. Is my reasoning correct in the first place?
real-analysis
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
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add a comment |
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I want to find for which values of $p$ the following series converges: $$sum_{n=1}^{infty}(-1)^n{{ln^pn}over{n}}$$
I found that for any value of $p$, $$lim_{nto infty} {ln^p nover n}=0$$
Now I have to prove that $f:x longmapsto{ln^p xover x}$ is a decreasing function to apply the alternating series test.
Its derivative is $$f'(x)={{pover x}ln^{p-1}x times x-ln^pxover x^2}={ln^{p-1}x(p-ln x)over x^2}$$ Thus, for $x>1,$ $$f'(x)leq0Leftrightarrow pleq ln{x}$$
And then I'm not quite sure whether it means something. Is my reasoning correct in the first place?
real-analysis
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
$endgroup$
$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
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– zhw.
Jan 20 at 19:37
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I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48
add a comment |
$begingroup$
I want to find for which values of $p$ the following series converges: $$sum_{n=1}^{infty}(-1)^n{{ln^pn}over{n}}$$
I found that for any value of $p$, $$lim_{nto infty} {ln^p nover n}=0$$
Now I have to prove that $f:x longmapsto{ln^p xover x}$ is a decreasing function to apply the alternating series test.
Its derivative is $$f'(x)={{pover x}ln^{p-1}x times x-ln^pxover x^2}={ln^{p-1}x(p-ln x)over x^2}$$ Thus, for $x>1,$ $$f'(x)leq0Leftrightarrow pleq ln{x}$$
And then I'm not quite sure whether it means something. Is my reasoning correct in the first place?
real-analysis
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
$endgroup$
I want to find for which values of $p$ the following series converges: $$sum_{n=1}^{infty}(-1)^n{{ln^pn}over{n}}$$
I found that for any value of $p$, $$lim_{nto infty} {ln^p nover n}=0$$
Now I have to prove that $f:x longmapsto{ln^p xover x}$ is a decreasing function to apply the alternating series test.
Its derivative is $$f'(x)={{pover x}ln^{p-1}x times x-ln^pxover x^2}={ln^{p-1}x(p-ln x)over x^2}$$ Thus, for $x>1,$ $$f'(x)leq0Leftrightarrow pleq ln{x}$$
And then I'm not quite sure whether it means something. Is my reasoning correct in the first place?
real-analysis
real-analysis
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
edited Jan 20 at 19:44
D. Appréhension
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
asked Jan 20 at 19:33
D. AppréhensionD. Appréhension
236
236
$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
$endgroup$
– zhw.
Jan 20 at 19:37
$begingroup$
I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48
add a comment |
$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
$endgroup$
– zhw.
Jan 20 at 19:37
$begingroup$
I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48
$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
$endgroup$
– zhw.
Jan 20 at 19:37
$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
$endgroup$
– zhw.
Jan 20 at 19:37
$begingroup$
I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48
$begingroup$
I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48
add a comment |
1 Answer
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So, if $n>mathrm e^p$, the general term $:dfrac{ln^pn}n:$ decreases to $0$, and Leibniz' criterion applies, since convergence of a series isn't modified if you change a finite number of terms.
answered Jan 20 at 19:47
BernardBernard
121k740116
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$begingroup$
So, if $n>mathrm e^p$, the general term $:dfrac{ln^pn}n:$ decreases to $0$, and Leibniz' criterion applies, since convergence of a series isn't modified if you change a finite number of terms.
answered Jan 20 at 19:47
BernardBernard
121k740116
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add a comment |
$begingroup$
So, if $n>mathrm e^p$, the general term $:dfrac{ln^pn}n:$ decreases to $0$, and Leibniz' criterion applies, since convergence of a series isn't modified if you change a finite number of terms.
answered Jan 20 at 19:47
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
So, if $n>mathrm e^p$, the general term $:dfrac{ln^pn}n:$ decreases to $0$, and Leibniz' criterion applies, since convergence of a series isn't modified if you change a finite number of terms.
answered Jan 20 at 19:47
BernardBernard
121k740116
$endgroup$
So, if $n>mathrm e^p$, the general term $:dfrac{ln^pn}n:$ decreases to $0$, and Leibniz' criterion applies, since convergence of a series isn't modified if you change a finite number of terms.
answered Jan 20 at 19:47
BernardBernard
121k740116
answered Jan 20 at 19:47
BernardBernard
121k740116
answered Jan 20 at 19:47
BernardBernard
121k740116
answered Jan 20 at 19:47
BernardBernard
121k740116
121k740116
add a comment |
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$begingroup$
You are moving in the right direction. You only need to show the terms are eventually decreasing. This means the terms decrease as soon as $ple ln n.$ (PS: $ln(n)^p$ isn't the right notation. )
$endgroup$
– zhw.
Jan 20 at 19:37
$begingroup$
I see, so can I say that f is decreasing for any value of $p$ because $ln x$ eventually becomes greater than $p$ at a certain point?
$endgroup$
– D. Appréhension
Jan 20 at 19:48