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I am trying to determine the behavior of the following series as $n\to\infty$. Let $0<\mu<1$ be fixed and for every positive integer $n\geq 1$, consider the function $f_n(t)$ of a real variable $t$ defined by the series $\sum_{k=0}^\infty\mu^k(1-\mu^kt)^n$. I want to determine how $f_n(t)$ behaves as $n\to\infty$ for $0<t<1$ (some kind of asymptotic formula).

Clearly $f_n(t)$ converges to $0$ for each $0<t<1$, but with what rate? And say, hypothetically, the rate is $O(1/n)$, then I would need to know at least what is $\lim_{n\to\infty}nf_n(t)$. I've tried several things for two weeks and I believe the rate of $O(1/n)$ is correct, but I can't find that limit. Any suggestions?

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Can you do the case $\mu = 1/2, t=1/2$ ? –  Gerald Edgar Apr 12 '11 at 0:34
    
I can't do the case $\mu=1/2,t=1/2$, or even $t=\mu$ –  Erwin Apr 12 '11 at 1:04
    
The limit does not exist in general. See my comment under "EDIT" below. –  GH from MO Apr 12 '11 at 21:57
    
@Erwin: I added more details for the second implication under "EDIT". –  GH from MO Apr 13 '11 at 23:37

4 Answers 4

up vote 9 down vote accepted

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 $$n=r^{2\ell} \ \Longrightarrow\ nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$ $$n=r^{2\ell+1}\ \Longrightarrow\ nf_n(t) \ll r^3 e^{-r} + r \ll r.$$ For the last estimate use that $\mu^x(1-\mu^xt)^n$ increases for $x\leq\ell-\frac{1}{2}$, hence $$ n\sum_{k=0}^{\ell-2}\mu^k(1-\mu^kt)^n < n\int_0^{\ell-1}\mu^x(1-\mu^xt)^n\ dx < r^2(1-r^{-2\ell})^{r^{2\ell+1}} < r^2 e^{-r}, $$ while clearly $$ n\sum_{k=\ell-1}^{\infty}\mu^k(1-\mu^kt)^n < n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n+\frac{n\mu^\ell}{1-\mu} < r^3 e^{-r}+2r.$$ Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.

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Why are you considering $nf_n(1)$ if you first fix $t=1/r^2$. The first implication is clear, the second I don't see, can you provide more details? –  Erwin Apr 13 '11 at 1:53
    
$f_n(1)$ was a typo, I fixed it. I gave more details for the second implication under "EDIT". –  GH from MO Apr 13 '11 at 23:37
    
I checked the details you provide proving that the limit in general does not exist and I am convinced now, clever argument, thanks a lot. I am kind of disappointed though. This problem was a water down version of a somewhat similar kind of series I was analyzing, I'll try to see now if perhaps in the original series something makes the limit to exist. I'll keep you updated if I find something interesting. –  Erwin Apr 14 '11 at 20:47

Using partial summation, we have that $$\sum_{k = 0}^{K}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - \mu^{K + 1})(1 - \mu^K t)^n}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{K}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx},$$ where $\lfloor x \rfloor$ is the integer part of $x$. By taking the limit as $K$ tends to infinity, $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx}.$$ Now a simple calculation shows that $$\frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^x t)^{n-1} \: dx} = \frac{(1 - t)^n}{1 - \mu}$$ by making the substitution $u = 1 - \mu^x t$. So the tricky part is the other bit of the integral, which is $$E = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{x + \lfloor x \rfloor} (1 - \mu^x t)^{n-1} \: dx}.$$ Note that as $x - 1 < \lfloor x \rfloor \leq x$, we have the bounds $$A \leq E \leq \frac{1}{\mu} A$$ with $$A = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{2x} (1 - \mu^x t)^{n-1} \: dx}.$$ Once again, this isn't tricky to calculate: making the same substitution as earlier, we find that $$A = \frac{\mu}{t (1 - \mu)} \frac{1 - (nt + 1) (1 - t)^n}{n + 1}.$$

So piecing everything together, we obtain $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - t)^n}{1 - \mu} + E$$ with $$E \asymp_{\mu} \frac{1 - (nt + 1) (1 - t)^n}{t(n + 1)}.$$ This doesn't yield a closed form for $\lim_{n \to \infty} n f_n(t)$, unfortunately, but it does show that $$\frac{\mu}{t(1 - \mu)} \leq \liminf_{n \to \infty} n f_n(t) \leq \limsup_{n \to \infty} n f_n(t) \leq \frac{1}{t (1 - \mu)}.$$

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@Peter Humphries: a guess there must be a sign mistake. The original sum was positive and your estimates are negative ... –  Fabian Apr 12 '11 at 9:27
    
Thanks Fabian, I think it's all correct now. –  Peter Humphries Apr 12 '11 at 10:32
    
@Peter: I have not followed every step in your argument, but you don't seem to prove that $\lim_{n\to\infty} nf_n(t)$ exists. All you prove is that $ntf_n(t)$ lies between two positive constants depending on $\mu$. This is in agreement with my finding above. However, I believe that $\lim_{n\to\infty} nf_n(t)$ does not exist: based on the diophantine properties of $n$ you can produce different limits along subsequences. –  GH from MO Apr 12 '11 at 15:17
    
More precisely, you give bounds for the $\liminf$ and the $\limsup$ that are similar to mine (in fact better than mine, but my argument can also be made effective in a similar fashion). –  GH from MO Apr 12 '11 at 15:20
    
Thanks very much for your responses. I had tried summation by parts but did not write the second summation as an integral, which leads to the nice estimates of Peter Humphries. I had also done some analysis in the spirit of GH though not as deep, that's how I came to believe that $O(1/n)$ was the right rate of decay. I have done other things and have some ideas that I will be posting tomorrow or today if I have a chance. If the limit $\lim_{n\to\infty}nf_n(t)$ does not exist, then the natural question is how does $nf_n(t)$ behave? Thanks again one more time. –  Erwin Apr 12 '11 at 21:59

Here is an elementary approach, which shows how to find the nature of $nf_n(t)$ as $n\to\infty$. But I'm not going to bound error terms or such so this remains an outline until those details are filled in.

Write $a(k) = \mu^k(1-t\mu^k)^n$ and consider $k$ as a real variable. $a(k)$ is easily seen to have a unique maximum at $k=k_0$ where $\mu^{-k_0}=1/((t(n+1))$. Now calculate $$a(k_0) = \frac{(n/(n+1))^n}{t(n+1)} \sim \frac{e^{-1}}{t(n+1)}$$ $$a(k_0+u) = a(k_0) \mu^u\left(\frac{1-\mu^u/(n+1)}{1-1/(n+1)}\right)^n \sim a(k_0) \mu^u \exp(1-\mu^u)),$$ the limits being for $n\to\infty$ with $u$ not too wild.

So now we have (modulo checking of error terms), $$nf_n(t) \sim \frac{1}{t} \sum_u \mu^u e^{-\mu^u},$$ where the sum is over $u\ge -k_0$ such that $k_0+u$ is an integer. The restriction $u\ge -k_0$ should be far enough in the tail that it doesn't matter, so we have $$nf_n(t) \sim \frac{1}{t} F_\mu(y),$$ where $y$ is the fractional part of $k_0$ and $$F_\mu(y) = \sum_{j=-\infty}^\infty \mu^{y+j}e^{-\mu^{y+j}}.$$ (Doesn't this last sum have a name? I'm sure I've seen it before.)

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As above, $\langle x\rangle$ denotes the fractional part of $x$.

The behavior of $f_n(t):=\sum_{k=0}^\infty\mu^k(1-\mu^k t)^n$ as $n\to\infty$ in the interval $(0,1)$ is as follows. For a subsequence $n_k$ of the natural numbers, the sequence $n_kf_{n_k}(t)$ converges if and only if $\langle-\log_\mu(n_k-1)\rangle$ converges as $k\to\infty$. Moreover, if $\langle-\log_\mu(n_k-1)\rangle\to q\in [0,1]$ as $k\to\infty$, then $\lim_{k\to\infty}n_kf_{n_k}(t)=F_q(t)$ uniformly on compacts of $(0,1)$, where $$ F_q(t)=\frac{\mu t}{1-\mu}\int_0^\infty\mu^{-\langle \log_\mu(s)+q\rangle}se^{-st}ds =\sum_{k=-\infty}^\infty\mu^{k-q}e^{t\mu^{k-q}}. $$

Since $\langle-\log_\mu n\rangle_{n=1}^\infty$ is dense in $[0,1]$, it follows that all the limit points of the sequence $n f_n(t)$ are precisely the functions $F_q(t)$, $0\leq q<1$. The same result is indeed valid for complex $t$ in the disk $|t-1/2|<1/2$.

The proof is a continuation of the ideas above of Humphries and GH.

We start with summation by parts $$ \sum_{k=0}^K\mu^k(1-\mu^kt)^n=\frac{(1-\mu^{K+1})(1-\mu^Kt)^n}{1-\mu}-\sum_{k=0}^{K-1}\frac{1-\mu^{k+1}}{1-\mu}[(1-\mu^{k+1}t)^n-(1-\mu^{k}t)^n]$$

$$=\frac{(1-t)^n}{1-\mu}-\frac{\mu^{K+1}(1-\mu^Kt)^n}{1-\mu}+\sum_{k=0}^{K-1}\frac{\mu^{k+1}}{1-\mu}[(1-\mu^{k+1}t)^n-(1-\mu^{k}t)^n]$$

Letting $K\to\infty$ we get

$$f_n(t)=O((1-t)^n)-\frac{\mu}{1-\mu}\sum_{k=0}^\infty\mu^{k}\int_{\mu^{k+1}}^{\mu^k}\frac{d}{dx}(1-xt)^ndx$$

$$=O((1-t)^n)+\frac{n\mu t}{1-\mu}\int_0^1\mu^{-\langle\log_\mu x\rangle}x(1-xt)^{n-1}dx$$

$$=O((1-t)^n)+\frac{n\mu t}{(1-\mu)(n-1)^2}\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}s(1-\frac{ts}{n-1})^{n-1}ds$$

Now, for $t\in(0,1)$, $s\in[0, n-1],$ we obviously have that

$$0\leq 1-\frac{ts}{n-1}\leq e^{-ts/(n-1)}$$ and that there is a constant $M$ independent of $n$ such that $$|e^{-ts/(n-1)}-(1-\frac{ts}{n-1})|\leq \frac{M s^2}{(n-1)^2}.$$

Hence $$|e^{-ts}-(1-\frac{ts}{n-1})^{n-1}|\leq |e^{-ts/(n-1)}-(1-\frac{ts}{n-1})|(n-1)e^{\frac{-ts(n-2)}{(n-1)}}$$ $$\leq \frac{M s^2e^{-ts/2}}{n-1},\quad n\geq 3,$$

and consequently,

$$\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}s(1-\frac{ts}{n-1})^{n-1}ds=\int_0^{n-1}\mu^{-\langle\log_\mu [s/(n-1)]\rangle}se^{-st}ds+O(1/n)$$

which together with the equality above yields

$$f_n(t)=\frac{n}{(n-1)^2}[\frac{\mu t}{1-\mu}\int_0^{\infty}\mu^{-\langle\log_\mu s-\log_\mu (n-1)\rangle}se^{-st}ds+O(1/n)]$$

and the claim made at the beginning follows.

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