Evaluate $\lim\limits_{n \to \infty} n \left( c(n)\ln\frac{c(n)}{0.5} +(1- c(n))\ln\frac{1-c(n)}{0.5}\right)$

Question

This is a follow up question on this question.

For $0<a,b<1$, evaluate $$\lim\limits_{n \to \infty} n \left( c(n)\log_2\frac{c(n)}{0.5} +(1- c(n))\log_2\frac{1-c(n)}{0.5}\right),$$ where $c(n)=\frac{0.5 a(\sqrt{n}-1)}{a(\sqrt{n}-1)+b}$.

I was thinking of using the Taylor expansion of $\ln(1-x)$ but it did not give me a any bounds.

Answer 1

Write $c=\frac{1+x}{2}$ so $x=\frac{b}{a\sqrt{n}}+o(\frac{1}{\sqrt{n}})$ and $$c\ln (2c)+(1-c)\ln (2(1-c))\approx\frac{1}{2}((1+x)(x-\tfrac{1}{2}x^2)+(1-x)(-x-\tfrac{1}{2}x^2))=\frac{1}{2}x^2+o(x^2).$$Hence you want $$\lim_{x\to 0^+}n\frac{1}{\ln 2}\frac{1}{2}x^2=\lim_{n\to\infty}\frac{n}{2\ln 2}\frac{b^2}{a^2n}=\frac{b^2}{a^2\ln 4}.$$

Answer 2

Let's plug in $a=0.5$ and $b=0.5$ then we have that $c=\frac{1}{2}\frac{\sqrt{n}-1}{\sqrt{n}}$. Then notice that the limit becomes

• $\lim_{n \to \infty} n[\frac{1}{2}\frac{\sqrt{n}-1}{\sqrt{n}} \ln \frac{\sqrt{n}-1}{\sqrt{n}}+ \frac{1}{2}\frac{\sqrt{n}+1}{\sqrt{n}} \log\frac{\sqrt{n}+1}{\sqrt{n}} ]$

which in turn is

• $\lim_{n \to \infty} [\frac{\sqrt{n}}{2}\sqrt{n}-1 \ln \frac{\sqrt{n}-1}{\sqrt{n}}+ \frac{\sqrt{n}}{2}\sqrt{n}+1 \log\frac{\sqrt{n}+1}{\sqrt{n}} ]$

which in turn is

• $\frac{1}{2 } \lim_{n \to \infty} [(n-\sqrt{n} )\ln \frac{\sqrt{n}-1}{\sqrt{n}}+( n+\sqrt{n}) \log\frac{\sqrt{n}+1}{\sqrt{n}} ]= -\infty $

Which clearly does not converge and your solution is incorrect you made approximations that were incorrect.

If you need further verification look at the solution that WolframAlpha gives when you plug in $a=0.5$ and $b=0.5$

https://www.wolframalpha.com/input/?i=lim+(n((0.5*0.5(sqrt(n))-1)%2F(0.5(sqrt(n)-1)%2B0.5))log(2*((0.5*0.5(sqrt(n))-1)%2F(0.5(sqrt(n)-1)%2B0.5)))-n(1-(0.5*0.5(sqrt(n))-1)%2F(0.5(sqrt(n)-1)%2B0.5))log(2-2*((0.5*0.5(sqrt(n))-1)%2F(0.5(sqrt(n)-1)%2B0.5))))

The computer does not lie.