10
$\begingroup$

For every $n\geq 3$, let $$ a_n=n\sum_{k=1}^\infty\frac{1}{4^{2k+1}}\left(1-\frac{1}{4^k} \right)^{n-2} $$ So for example, $$ a_3=3\sum_{k=1}^\infty 2^{-4k-2}-2^{-6k-2}=\frac{3}{4}\left(\frac{1}{15}-\frac{1}{63}\right) $$ I am curious if $a_n$ converges as $n\rightarrow\infty$, and if so, to what limit. One thing I've tried is to simplify the terms using the binomial expansion: $$\begin{align} a_n&=n\sum_{k=1}^\infty2^{-2(2k+1)}\sum_{m=0}^{n-2}{n-2\choose m}(-2^{-2k})^m\\&=\frac{n}{4}\sum_{m=0}^{n-2}{n-2\choose m}(-1)^m\sum_{k=1}^\infty 2^{-(2m+4)k}\\&=\frac{n}{4}\sum_{m=0}^{n-2}{n-2\choose m}\frac{(-1)^m}{4^{m+2}-1}\end{align} $$ But I don't know how to simplify it further and to evaluate whether or not it converges. Any hint would be greatly appreciated.

$\endgroup$
3
  • $\begingroup$ I get $\sim\frac{\ln n}{n^2}$ @Gary $\endgroup$
    – MathFail
    Commented May 30, 2023 at 13:47
  • $\begingroup$ @MathFail I did some new calculations using a Stieltjes integral and found that $a_n$ decays like $n^{-1}$. Of course your upper bound is enought to conclude $a_n\to 0$. $\endgroup$
    – Gary
    Commented May 31, 2023 at 1:12
  • $\begingroup$ Replacing the sum by an integral gives $a(n) = n \int_1^{\infty } \frac{\left(1-\frac{1}{4^k}\right)^{n-2}}{4^{2 k+1}} \, dk = \frac{4^{-n-1} \left(3\ 4^n-3^n (n+3)\right)}{3 (n-1) \log (4)} \simeq \frac{1}{4 n \log (4)}\to 0$. $\endgroup$ Commented May 31, 2023 at 19:52

2 Answers 2

5
$\begingroup$

Let $$f(k)=\frac{1}{4^{2k+1}}\left(1-\frac{1}{4^k} \right)^{n-2}$$

For a fixed $n$, take derivative $f'(k)=0$, the maxima occurs at

$$4^{k_M}=\frac{n}2,~~~~~~f(k_M)=\frac{1}{n^2}\left( 1-\frac2 n\right)^{n-2}$$

so we split the sum into two parts:

$$\begin{align}\sum_{k=1}^{\infty}f(k)&=\sum_{k=1}^{\lceil k_M\rceil}f(k)+\sum_{k=\lceil k_M\rceil}^{\infty}f(k)\\ \\ &\le f(k_M)\lceil k_M\rceil+\sum_{k=\lceil k_M\rceil}^{\infty}\frac{1}{4^{2k+1}}\\ \\ &\le \frac{\log_4(\frac n2)}{n^2}\left( 1-\frac2 n\right)^{n-2}+\frac{16}{15n^2} \end{align}$$ therefore,

$$0\le a_n\le \frac{\log_4(\frac n2)}{n}\left( 1-\frac2 n\right)^{n-2}+\frac{16}{15n} $$ By squeeze theorem,

$$\lim a_n=0$$

$\endgroup$
1
  • $\begingroup$ amazing! thanks so much! $\endgroup$
    – Y.Z.
    Commented Jun 2, 2023 at 6:16
0
$\begingroup$

Not a complete solution (yet) but two ideas

  1. Replace the sum by an integral (as suggested in my comment). This gives

$$\begin{align}a(n) &\to i(n):=n \int_{1}^{\infty} \frac{1}{4^{2k+1}}\left(1-\frac{1}{4^k}\right)^{n-2}\;dk\\&=\frac{4^n -3^{n-1}(n+3)}{(n-1) 4^{n+1}\log(4)}\\&\to -\frac{(\frac{3}{4})^n}{12 \log(4)}+\frac{1}{4n \log(4)}\end{align}\tag{0}$$

This goes to $0$ for $n\to\infty$. But the question is how to justify the replacement, especially in the present case where the integrand/summand is not monotonous as a function of $k$.

  1. Form the generating function of $a_n$, where

$$a_n = i(n) = n\sum_{k=1}^\infty\frac{1}{4^{2k+1}}\left(1-\frac{1}{4^k} \right)^{n-2} \tag{1}$$

and use the reasoning of this reference to find the asymptotic form of $a_n$ for $n \to \infty$.

Here we go.

Letting $z=\frac{1}{4}$ the generating function is given by

$$g(t,z) = \sum_{n=2}^{\infty}a_n t^n = \frac{1}{4}\sum_{n=2}^{\infty}\sum_{k=1}^\infty n \;t^n z^{2k}\left(1-z^{k} \right)^{n-2} \tag{2}$$

It can be written as $$g(t,z) = \frac{1}{4} t \frac {d}{dt} h(t,z)\tag{3}$$

where

$$\begin{align}h(t,z)& =\sum_{n=2}^{\infty}\sum_{k=1}^\infty \;t^n z^{2k}\left(1-z^{k} \right)^{n-2}\\ &=\sum_{k=1}^\infty \sum_{n=2}^{\infty} t^n z^{2k}\left(1-z^{k} \right)^{n-2}\\ &=\sum_{k=1}^\infty\frac{t^2 z^{2 k}}{1-t \left(1-z^k\right)} \end{align}\tag{4}$$

Whence follows

$$g(z,t) = \frac{1}{4} \sum_{k=1}^{\infty}b(t,k)\tag{5a}$$

where

$$b(t,k)=\frac{t^2 \left(4^{-k} t-t+2\right)}{4^{2 k} \left(4^{-k} t-t+1\right)^2}\tag{5b}$$

Now we wish to replace the $k$-sum somehow by an integral. To this end we observe that for given $t \in (0,1)$ the summand $b(t,k)$ is a positive monotonously decreasing function of $k$ so that from graphic considerations we can write down the inequality

$$b(t,1)+\int_{1}^{\infty}b(t,k)\;dk>\sum_{k\ge 1}b(t,k)>\int_{1}^{\infty}b(t,k)\;dk \tag{6}$$

And we have

$$b(t,1) =1 + 4^{-t}\tag{7a}$$

$$I(t)=\int_{1}^{\infty}b(t,k)\;dk = \frac{1}{8 \log (2)}\left(\frac{3 t^2}{3 t-4}+4 t (\log (4-3 t)-\log (4-4 t))\right)\tag{7b}$$

I can see that $I(t)$ has a simple pole at $t=\frac{4}{3}$ and a branch point in the same location, furthermore there is a branch point at $t=1$.

Close to each of these singularities our sum $g$ is squeezed to the values of $I$ by means of the inequality $(6)$.

But here I am stuck and ask for help.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .