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.

  • $\begingroup$ 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

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) $.

  • $\begingroup$ 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
  • 1
    $\begingroup$ I think in the outer sum there should be a $2^{-m}$, not $2^{m}$. $\endgroup$ – Mah Apr 23 '18 at 3:23

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 $n$ $$y^n <1+t+\frac 12 t^2+\frac 16 t^3\tag 2$$

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.