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I am trying to find a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ as a function of $x$ and $n$ and for large $n$, where $x$ changes with $n$ such that $\lim_{n\to\infty}x=0$.

I am not sure wether my approach to solve this is right or not, but this is what I did: \begin{align*}\left(\frac{1+x}{(1+x/2)^2}\right)^n&=e^{n(\ln({1+x})-2\ln{(1+x/2)})}\\ &=e^{n(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots-2(\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{24}-\cdots))}\\ &=e^{n(-\frac{x^2}{4}+\frac{x^3}{4}-\frac{15x^4}{64}+\cdots)}\\ &\geq e^{n(-\frac{x^2}{4})}\\ &=1-(n\frac{x^2}{4})+(n\frac{x^2}{4})^2-\cdots \\ &\geq 1-(n\frac{x^2}{4}) \end{align*} We know $\lim_{n\to\infty}x=0$, but we don't know whether $\lim_{n\to\infty}nx^2=0$ . Hence, the last inequality is not necessarily correct, because the sum of the terms after $1-(n\frac{x^2}{4}) $ may not be greater than zero.

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  • $\begingroup$ how is $x$ a function of $n$??? $\endgroup$ – tired Apr 16 '18 at 22:58
  • $\begingroup$ The exact relationship between $x$ and $n$ is not specified. We only know $\lim_{n\to\infty}x=0$. $\endgroup$ – Mah Apr 17 '18 at 0:09
  • $\begingroup$ For $n$ large enough you could use Bernoulli's inequality to bound $(1 + x)^n$ from below by $1 + nx$. From there I'm not sure how to proceed without more information about $x(n)$. (This probably isn't a "tight" bound anyway.) $\endgroup$ – rwbogl Apr 17 '18 at 3:51
  • $\begingroup$ @rwbogl and how should I bound the term in the denominator? $\endgroup$ – Mah Apr 17 '18 at 4:32
  • $\begingroup$ I seem to get $e^{-nx^2/4}$ (I think your $16$ should be an $8$) but the difference hardly matters: you clearly get different limits with $x=n^{-1}$, $x=n^{-1/2}$ and $x=n^{-1/4}$ $\endgroup$ – Henry Apr 17 '18 at 9:42
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Write $y = x/2$. Then $\left(\frac{1+x}{(1+x/2)^2}\right)^n =\left(\frac{1+2y}{(1+y)^2}\right)^n =\frac{(1+2y)^n}{(1+y)^{2n}} $.

Since $(1+2y)^n =\sum_{j=0}^n \binom{n}{j}2^jy^j $ and $\frac1{(1+y)^{2n}} =\sum_{k=0}^{\infty} \binom{2n+k-1}{k}(-1)^ky^k $,

$\begin{array}\\ \frac{(1+2y)^n}{(1+y)^{2n}} &=\sum_{j=0}^n \binom{n}{j}2^jy^j\sum_{k=0}^{\infty} \binom{2n+k-1}{k}(-1)^ky^k\\ &=\sum_{j=0}^n\sum_{k=0}^{\infty}y^{j+k} \binom{n}{j}2^j \binom{2n+k-1}{k}(-1)^k\\ &=\sum_{m=0}^{\infty}y^m\sum_{j=0}^n\binom{n}{j}2^j \binom{2n+m-j-1}{m-j}(-1)^{m-j}\qquad j+k = m, k = m-j\\ &=\sum_{m=0}^{\infty}y^m(-1)^m\sum_{j=0}^n\dfrac{n!(2n+m-j-1)!}{j!(n-j)!(m-j)!(2n-1)!}2^j (-1)^{j}\\ &=\dfrac{n!}{(2n-1)!}\sum_{m=0}^{\infty}y^m(-1)^m\sum_{j=0}^n\dfrac{(2n+m-j-1)!}{j!(n-j)!(m-j)!}2^j (-1)^{j}\\ \text{so}\\ \frac{(1+x)^n}{(1+x/2)^{2n}} &=\dfrac{n!}{(2n-1)!}\sum_{m=0}^{\infty}(-1)^m2^{-m}x^m\sum_{j=0}^{\min(m, n)}\dfrac{(2n+m-j-1)!}{j!(n-j)!(m-j)!}2^j (-1)^{j}\\ \end{array} $

With this, you can get the power series.

Note: Wolfy says this starts like

$1-\dfrac{nx^2}{4}+\dfrac{nx^3}{4}+\dfrac{n(n-7)x^4}{32} -\dfrac{n(n - 3) x^5}{16} - \dfrac{n (n^2 - 33 n + 62) x^6}{384}+O(x^7) $.

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  • $\begingroup$ Thanks a lot! this is exactly what I was looking for. I just don't understand how $n$ turned into $\min(m,n)$ (in the upper limit of the inner summation)? $\endgroup$ – Mah Apr 19 '18 at 14:01
  • $\begingroup$ Because of the $(n-j)!(m-j)!$ in the denominator. $\endgroup$ – marty cohen Apr 19 '18 at 15:08
  • $\begingroup$ I see. So that is because $j+k=m$ and thus $j\leq m$. $\endgroup$ – Mah Apr 19 '18 at 18:26
  • $\begingroup$ That is correct. $\endgroup$ – marty cohen Apr 19 '18 at 20:14
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    $\begingroup$ I think in the outer sum there should be a $2^{-m}$, not $2^{m}$. $\endgroup$ – Mah Apr 23 '18 at 3:23
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Start considering $$y=\frac{1+x}{1+\left(\frac x2 \right)^2}$$ and use Taylor around $x=0$; this will give $$y=1+x-\frac{x^2}{4}-\frac{x^3}{4}+O\left(x^4\right)$$ Now, use the binomial expansion and get $$y^n=1+nx+\frac n4 (2n-3)x^2+\frac n{12}(2 n^2-9 n+4)x^3+O\left(x^4\right)$$ Define $t=nx$ and you can write as $$y^n=1+t+\left(\frac{1}{2}-\frac{3}{4 n}\right) t^2+\left(\frac{1}{6}-\frac{3}{4 n}+\frac{1}{3 n^2}\right)t^3+O\left(t^5\right)\tag 1$$ So, for large $n$ $$y^n <1+t+\frac 12 t^2+\frac 16 t^3\tag 2$$

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    $\begingroup$ I think the question says $\left(1+\frac x2 \right)^2$ rather than $1+\left(\frac x2 \right)^2$ $\endgroup$ – Henry Apr 17 '18 at 9:43

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