Mixing
For which $p$ does $sumlimits_{n=1}^{infty}left(1-frac{p ln(n)}{n}right)^{n}$ converge?
$begingroup$
$$sumlimits_{n=1}^{infty}left(1-frac{p ln(n)}{n}right)^{n}$$
For which $p$ does it converge?
I tried to write it in a different way :
$$left(1-frac{p ln(n)}{n}right)^{n} = left[left(1-frac{1}{frac{n}{pln(n)}}right)^{frac{n}{pln(n)}}right]^{-pln(n)} sim e^{-pln(n)}$$
so i have to compare the original one with $e^{-pln(n)}$ which is convergent for $p>1$.
I am not sure if that is correct and how to prove that the lim exist with the comparison test.
sequences-and-series convergence
edited Jan 6 at 20:44
rtybase
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
$endgroup$
add a comment |
$begingroup$
$$sumlimits_{n=1}^{infty}left(1-frac{p ln(n)}{n}right)^{n}$$
For which $p$ does it converge?
I tried to write it in a different way :
$$left(1-frac{p ln(n)}{n}right)^{n} = left[left(1-frac{1}{frac{n}{pln(n)}}right)^{frac{n}{pln(n)}}right]^{-pln(n)} sim e^{-pln(n)}$$
so i have to compare the original one with $e^{-pln(n)}$ which is convergent for $p>1$.
I am not sure if that is correct and how to prove that the lim exist with the comparison test.
sequences-and-series convergence
edited Jan 6 at 20:44
rtybase
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
$endgroup$
$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54
add a comment |
$begingroup$
$$sumlimits_{n=1}^{infty}left(1-frac{p ln(n)}{n}right)^{n}$$
For which $p$ does it converge?
I tried to write it in a different way :
$$left(1-frac{p ln(n)}{n}right)^{n} = left[left(1-frac{1}{frac{n}{pln(n)}}right)^{frac{n}{pln(n)}}right]^{-pln(n)} sim e^{-pln(n)}$$
so i have to compare the original one with $e^{-pln(n)}$ which is convergent for $p>1$.
I am not sure if that is correct and how to prove that the lim exist with the comparison test.
sequences-and-series convergence
edited Jan 6 at 20:44
rtybase
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
$endgroup$
$$sumlimits_{n=1}^{infty}left(1-frac{p ln(n)}{n}right)^{n}$$
For which $p$ does it converge?
I tried to write it in a different way :
$$left(1-frac{p ln(n)}{n}right)^{n} = left[left(1-frac{1}{frac{n}{pln(n)}}right)^{frac{n}{pln(n)}}right]^{-pln(n)} sim e^{-pln(n)}$$
so i have to compare the original one with $e^{-pln(n)}$ which is convergent for $p>1$.
I am not sure if that is correct and how to prove that the lim exist with the comparison test.
sequences-and-series convergence
sequences-and-series convergence
edited Jan 6 at 20:44
rtybase
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
edited Jan 6 at 20:44
rtybase
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
edited Jan 6 at 20:44
rtybase
10.7k21533
edited Jan 6 at 20:44
rtybase
10.7k21533
edited Jan 6 at 20:44
rtybase
10.7k21533
10.7k21533
asked Jan 6 at 19:43
Mather Mather
3047
asked Jan 6 at 19:43
Mather Mather
3047
asked Jan 6 at 19:43
Mather Mather
3047
3047
$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54
add a comment |
$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54
$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54
$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By the Taylor series expansion of $x mapsto ln(1-x)$, $0le x<1$, near $0$, one has
$$
ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n to infty$,
begin{align}
left(1-frac{p ln(n)}{n}right)^{n}&=e^{nln left(1-frac{p ln(n)}{n} right)}
\\&=e^{-nfrac{p ln(n)}{n} +n,Oleft(frac{ln^2 n}{n^2}right)}
\\&=e^{ln frac{1}{n^p}}times e^{Oleft(frac{ln^2 n}nright)}
\\&=frac{1}{n^p}times e^{o(1)}
\\&sim frac{1}{n^p}
end{align}
the given series converges iff $p>1$ by the limit comparison test.
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
add a comment |
$begingroup$
Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative.
However, if $p>e$, $left(1-frac{pln(n)}{n}right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $exists n_p in mathbb{N}$ such that $forall n > n_p, left(1-frac{pln(n)}{n}right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
By the Taylor series expansion of $x mapsto ln(1-x)$, $0le x<1$, near $0$, one has
$$
ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n to infty$,
begin{align}
left(1-frac{p ln(n)}{n}right)^{n}&=e^{nln left(1-frac{p ln(n)}{n} right)}
\\&=e^{-nfrac{p ln(n)}{n} +n,Oleft(frac{ln^2 n}{n^2}right)}
\\&=e^{ln frac{1}{n^p}}times e^{Oleft(frac{ln^2 n}nright)}
\\&=frac{1}{n^p}times e^{o(1)}
\\&sim frac{1}{n^p}
end{align}
the given series converges iff $p>1$ by the limit comparison test.
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
add a comment |
$begingroup$
By the Taylor series expansion of $x mapsto ln(1-x)$, $0le x<1$, near $0$, one has
$$
ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n to infty$,
begin{align}
left(1-frac{p ln(n)}{n}right)^{n}&=e^{nln left(1-frac{p ln(n)}{n} right)}
\\&=e^{-nfrac{p ln(n)}{n} +n,Oleft(frac{ln^2 n}{n^2}right)}
\\&=e^{ln frac{1}{n^p}}times e^{Oleft(frac{ln^2 n}nright)}
\\&=frac{1}{n^p}times e^{o(1)}
\\&sim frac{1}{n^p}
end{align}
the given series converges iff $p>1$ by the limit comparison test.
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
add a comment |
$begingroup$
By the Taylor series expansion of $x mapsto ln(1-x)$, $0le x<1$, near $0$, one has
$$
ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n to infty$,
begin{align}
left(1-frac{p ln(n)}{n}right)^{n}&=e^{nln left(1-frac{p ln(n)}{n} right)}
\\&=e^{-nfrac{p ln(n)}{n} +n,Oleft(frac{ln^2 n}{n^2}right)}
\\&=e^{ln frac{1}{n^p}}times e^{Oleft(frac{ln^2 n}nright)}
\\&=frac{1}{n^p}times e^{o(1)}
\\&sim frac{1}{n^p}
end{align}
the given series converges iff $p>1$ by the limit comparison test.
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
By the Taylor series expansion of $x mapsto ln(1-x)$, $0le x<1$, near $0$, one has
$$
ln(1-x)=-x+O(x^2).
$$
Then one may write, as $n to infty$,
begin{align}
left(1-frac{p ln(n)}{n}right)^{n}&=e^{nln left(1-frac{p ln(n)}{n} right)}
\\&=e^{-nfrac{p ln(n)}{n} +n,Oleft(frac{ln^2 n}{n^2}right)}
\\&=e^{ln frac{1}{n^p}}times e^{Oleft(frac{ln^2 n}nright)}
\\&=frac{1}{n^p}times e^{o(1)}
\\&sim frac{1}{n^p}
end{align}
the given series converges iff $p>1$ by the limit comparison test.
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
edited Jan 7 at 20:38
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
answered Jan 6 at 19:51
Olivier OloaOlivier Oloa
108k17177294
108k17177294
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
add a comment |
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
The limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– Olivier Oloa
Jan 6 at 20:00
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
that is perfect but the thing is i have to show formally that the limit of the comparision exists and it is greater than $0$ and less than $infty$
$endgroup$
– Mather
Jan 6 at 20:09
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
$begingroup$
Actually, the limit comparison test alone is not sufficient without some pre-work. See my answer below
$endgroup$
– Harmonic Sun
Jan 6 at 20:12
1
1
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
$begingroup$
@Mather Since $Oleft(frac{ln^2 n}nright) to 0$ as $n to infty$, then $e^{Oleft(frac{ln^2 n}nright)} to e^0=1$ thus $frac{u_n}{v_n} to 1$ as $n to infty$ where $u_n=left(1-frac{p ln(n)}{n}right)^{n}$, $v_n=frac{1}{n^p}$ and the limit comparison test applies.
$endgroup$
– Olivier Oloa
Jan 7 at 5:46
add a comment |
$begingroup$
Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative.
However, if $p>e$, $left(1-frac{pln(n)}{n}right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $exists n_p in mathbb{N}$ such that $forall n > n_p, left(1-frac{pln(n)}{n}right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
$endgroup$
add a comment |
$begingroup$
Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative.
However, if $p>e$, $left(1-frac{pln(n)}{n}right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $exists n_p in mathbb{N}$ such that $forall n > n_p, left(1-frac{pln(n)}{n}right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
$endgroup$
add a comment |
$begingroup$
Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative.
However, if $p>e$, $left(1-frac{pln(n)}{n}right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $exists n_p in mathbb{N}$ such that $forall n > n_p, left(1-frac{pln(n)}{n}right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
$endgroup$
Both the OP's suggestion and Olivier Oloa's answer use the limit comparison test.
Actually, it's not that simple. The limit comparison test requires 2 sequences that are either both always positive, or both always negative.
However, if $p>e$, $left(1-frac{pln(n)}{n}right)^n$ is not always positive.
Now, one can show pretty easily that for every $p$, $exists n_p in mathbb{N}$ such that $forall n > n_p, left(1-frac{pln(n)}{n}right)^n>0$ (that is to say, that although this sequence is not always positive, there is a certain point from where it's always positive).
And then we can apply the limit comparison test to sequence new sequence optained by starting at this $n_p$ instead of $n=1$. And since this is a limit test (asymptotic by nature), an offset of $n_p$ doesn't change the result.
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
answered Jan 6 at 20:11
Harmonic SunHarmonic Sun
5618
5618
add a comment |
add a comment |
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$begingroup$
Your interesting path can be made more rigorous, please see my answer below.
$endgroup$
– Olivier Oloa
Jan 6 at 19:54