|
15
|
|
edited Apr 13 2011 at 23:46 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] $$
$$ = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>1$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 2^{-re^{-r} + r \ll r.$$
For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence
$$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r}e^{-r}, $$
while clearly
$$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r.$$
e^{-r}+2r.$$
Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
14
|
|
edited Apr 13 2011 at 23:41 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] $$
$$ = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>1$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 2^{-r} + r \ll r.$$
For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-1$ and decreases for $x\geq\ell$, x\leq\ell-\frac{1}{2}$, hence
$$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r} 2^{-r}, $$
and also
while clearly
$$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r.$$
Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
13
|
|
edited Apr 13 2011 at 23:32 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] $$
$$ = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>0$ r>1$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll \max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + r \ll r.$$
For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-1$ and decreases for $x\geq\ell$, hence
$$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 2^{-r} $$
and also
$$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 2^{-r}+2r.$$
Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.
|
|
|
|
12
|
|
edited Apr 13 2011 at 4:28 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>0$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf_n(1nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(1nf_n(t) \ll \max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + r \ll r.$$
Hence $\lim_{n\to\infty} nf_n(1)$ nf_n(t)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
11
|
|
edited Apr 12 2011 at 22:15 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT:I think it It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=1$ t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$ \mu:=\frac{1}{r^2}$, where $r>0$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\ nf_n(1) > (1-1/n)^n n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg 1,$$
r^2,$$
$$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(1) \ll r2^{-r\max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + 1/r r \ll 1/r.$$
r.$$
Hence $\lim_{n\to\infty} nf_n(1)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
10
|
|
edited Apr 12 2011 at 22:02 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: I think it is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=1$ and $\mu:=\frac{1}{r^2}$ where $r>0$ is an integer. Then for any integer $k>0$ \ell>0$ we have
$$n=r^{2k} $n=r^{2\ell} \ Longrightarrow \Longrightarrow\ nf_n(1) > (1-1/n)^n \gg 1,$$
$$n=r^{2k+1} $n=r^{2\ell+1}\ \Longrightarrow Longrightarrow\ nf_n(1) \ll r2^{-r} + 1/r \ll 1/r.$$
Hence $\lim_{n\to\infty} nf_n(1)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
9
|
|
edited Apr 12 2011 at 21:56 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: I think it is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=1$ and $\mu:=\frac{1}{r^2}$ where $r>0$ is an integer. Then for any integer $k>0$ we have
$$n=r^{2k} \Longrightarrow nf_n(1) > (1-1/n)^n \gg 1,$$
$$n=r^{2k+1} \Longrightarrow nf_n(1) \ll r2^{-r} + 1/r \ll 1/r.$$
Hence $\lim_{n\to\infty} nf_n(1)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
8
|
|
edited Apr 12 2011 at 21:50 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
EDIT: I think it is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=1$ and $\mu:=\frac{1}{r^2}$ where $r>0$ is an integer. Then
$$n=r^{2k} \Longrightarrow nf_n(1) > (1-1/n)^n \gg 1,$$
$$n=r^{2k+1} \Longrightarrow nf_n(1) \ll r2^{-r} + 1/r \ll 1/r.$$
Hence $\lim_{n\to\infty} nf_n(1)$ does not exist when $r$ is sufficiently large.
|
|
|
|
|
7
|
|
edited Apr 12 2011 at 14:58 |
This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] =
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
In the last integral the integrand Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence it the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.
Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.
|
|
|
|
|
6
|
|
edited Apr 12 2011 at 14:35 |
My first This is my third responsewas wrong, so let me correct it. I claim now that indeedfor $0 < t < 1$ we have the uniform bounds$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$$$ For the lower bound observe \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$where $O(1)$ denotes quantities that the largest term occurs for are absolutely bounded. First of all,$\mu^k\asymp_\mu $ f_n(t) = \frac{1}{nt}$ and it contributes int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] = \int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$where $\asymp_\mu\frac{1}{nt}$ to x=[x]+\langle x\rangle$ is the sumdecomposition into integral and fractional parts. For Evaluating the upper bound observe that first integral on the function $x\mapsto \mu^x (1-\mu^x t)^n$ increases when $\mu^x>\frac{1}{(n+1)t}$, decreases when $\mu^x<\frac{1}{(n+1)t}$, and has a maximum $\asymp\frac{1}{nt}$ when $\mu^x=\frac{1}{(n+1)t}$. Thereforeright, by and applying integration parts on the familiar technique of comparing an infinite sum to an second integral, we obtain$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n \ll \frac{1}{nt}+\int_0^\infty f_n(t) = \mu^x frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-\mu^x t)^n\ dx 1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \ll_\mu langle x\rangle\ frac{1}{nt}.$$Note that dx.$$In the last integral the integrand changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence it can be evaluated explicitly estimated readily by $0\leq \langle x\rangle < 1$ as$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$It follows that$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$which implies the change of variable claim above, noting that $s:=\mu^x$, say.\lim_{n\to\infty}n(1-t)^{n}=0$. Based on this the above argument , I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. This is because for the integral above the limit does existMore precisely, but the difference between the sum and I don't think that$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$exists, because the integral is very sensitive to the nearest value of $\mu^k$ to $\frac{1}{nt}$. In other words, the fine behavior of $n f_n(t)$ depends on the fractional part of $\log_\mu(nt)$. EDIT: I fixed \log_\mu\frac{1}{(n+1)t}$ where the maximum in $x$.integrand changes its sign.
|
|
|
|
|
5
|
|
edited Apr 12 2011 at 11:16 |
My first response was wrong, so let me correct it. I claim now that indeed
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. $$
For the lower bound observe that the largest term occurs for $\mu^k\asymp_\mu \frac{1}{nt}$ and it contributes $\asymp_\mu\frac{1}{nt}$ to the sum. For the upper bound observe that the function $x\mapsto \mu^x (1-\mu^x t)^n$ increases when $\mu^x>\frac{1}{nt}$, \mu^x>\frac{1}{(n+1)t}$, decreases when $\mu^x<\frac{1}{nt}$, <\frac{1}{(n+1)t}$, and has a maximum $\asymp\frac{1}{nt}$ when $\mu^x=\frac{1}{nt}$. \mu^x=\frac{1}{(n+1)t}$. Therefore, by the familiar technique of comparing an infinite sum to an integral, we obtain
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n \ll \frac{1}{nt}+\int_0^\infty \mu^x (1-\mu^x t)^n\ dx \ \ll_\mu \frac{1}{nt}.$$
Note that the integral can be evaluated explicitly by the change of variable $s:=\mu^x$, say.
Based on this argument, I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. This is because for the integral above the limit does exist, but the difference between the sum and the integral is very sensitive to the nearest value of $\mu^k$ to $\frac{1}{nt}$. In other words, the fine behavior of $n f_n(t)$ depends on the fractional part of $\log_\mu(nt)$.
EDIT: I fixed the maximum in $x$.
|
|
|
|
|
4
|
|
edited Apr 12 2011 at 6:10 |
My first response was wrong, so let me correct it. I claim now that indeed
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. $$
For the lower bound observe that the largest term occurs for $\mu^k\asymp_\mu \frac{1}{nt}$ and it contributes $\asymp_\mu\frac{1}{nt}$ to the sum. For the upper bound observe that the function $x\mapsto \mu^x (1-\mu^x t)^n$ increases when $\mu^x>\frac{1}{nt}$, decreases when $\mu^x<\frac{1}{nt}$, and has a maximum $\asymp\frac{1}{nt}$ when $\mu^x=\frac{1}{nt}$. Therefore, by the familiar technique of comparing an infinite sum to an integral, we obtain
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n \ll \frac{1}{nt}+\int_0^\infty \mu^x (1-\mu^x t)^n\ dx \ \ll_\mu \frac{1}{nt}.$$
Note that the integral can be evaluated explicitly by the change of variable $s:=\mu^x$, say.
Based on this argument, I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. This is because for the integral above the limit does exist, but the difference between the sum and the integral is very sensitive to the nearest value of $\mu^k$ to $\frac{1}{nt}$. In other words, the fine behavior of $n f_n(t)$ depends on the fractional part of $\log_\mu(nt)$.
|
|
|
|
|
3
|
|
edited Apr 12 2011 at 5:59 |
My first response was wrong, so let me correct it. I claim now that indeed
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. $$
For the lower bound observe that any the largest term with occurs for $\mu^k\asymp_\mu \frac{1}{nt}$ and it contributes $\asymp_\mu\frac{1}{nt}$ to the sum. For the upper bound estimate observe that the sum with a related integral function $x\mapsto \mu^x (using monotonicity properties of 1-\mu^x t)^n$ increases when $\mu^x>\frac{1}{nt}$, decreases when $\mu^x<\frac{1}{nt}$, and has a maximum $\asymp\frac{1}{nt}$ when $\mu^x=\frac{1}{nt}$. Therefore, by the terms):
familiar technique of comparing an infinite sum to an integral, we obtain
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n < \sum_{k=0}^\infty\mu^k e^{-nt\mu^k} ll \ll\frac{1}{nt}+\int_0^\infty frac{1}{nt}+\int_0^\infty \mu^x e^{-nt\mu^x}(1-\mu^x t)^n\ dx \ \ll_\mu \frac{1}{nt}.$$
Note that the integral can be evaluated explicitly by the change of variable $s:=\mu^x$, say.
|
|
|
|
|
2
|
|
edited Apr 12 2011 at 5:32 |
My first response was wrong, so let me correct it. I claim now that indeed
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{\log(nt)}{nt}. sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. $$
For the lower bound observe that $\frac{1}{nt}<\mu^k<\frac{2}{nt}$ holds in a range $k\asymp_\mu\log(nt)$ and any such term with $k$ \mu^k\asymp_\mu \frac{1}{nt}$ contributes $\gg\frac{1}{nt}$ \asymp_\mu\frac{1}{nt}$ to the sum. For the upper bound split and estimate the sum as followswith a related integral (using monotonicity properties of the terms):
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n = < \sum_{k:\mu^k<\lambda}\dots + sum_{k=0}^\infty\mu^k e^{-nt\mu^k} \sum_{k:\mu^k\geq\lambda}\dots < ll\frac{1}{nt}+\int_0^\infty \frac{\lambda}{1-\mu}+\frac{(1-\lambda t)^n}{1-\mu}\ll_\mu mu^x e^{-nt\mu^x}\ lambda+e^{-\lambda nt}.$$
For $\lambda:=\frac{\log(nt)}{nt}$ the right hand side is $\ll\frac{\log(nt)}{nt}$, so we are done.dx \ \ll_\mu \frac{1}{nt}.$$
Note : The above bounds show that the $O(1/n)$ conjecture fails for any integral can be evaluated explicitly by the change of variable $0 < t < 1$s:=\mu^x$, say.
|
|
|
|
|
1
|
|
answered Apr 12 2011 at 3:06 |
I claim that
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{\log(nt)}{nt}. $$
For the lower bound observe that $\frac{1}{nt}<\mu^k<\frac{2}{nt}$ holds in a range $k\asymp_\mu\log(nt)$ and any such $k$ contributes $\gg\frac{1}{nt}$ to the sum. For the upper bound split and estimate the sum as follows:
$$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n = \sum_{k:\mu^k<\lambda}\dots + \sum_{k:\mu^k\geq\lambda}\dots < \frac{\lambda}{1-\mu}+\frac{(1-\lambda t)^n}{1-\mu}\ll_\mu \lambda+e^{-\lambda nt}.$$
For $\lambda:=\frac{\log(nt)}{nt}$ the right hand side is $\ll\frac{\log(nt)}{nt}$, so we are done.
Note: The above bounds show that the $O(1/n)$ conjecture fails for any $0 < t < 1$.
|
|
|
