Proof the limit tend to 0(solution verification)

Question

Given

$h_{n+1}=\surd (h_n)^2-b(h_{n})^{2.1}$

$h_0=8$

$0.8<b<0$

I want to prove the limit tend to $0$ for all values of b in between 0.8 and 0 when n tend to infinity so I plot the graph using script below

function curve1(t,b,h,a)

i=1

while a<=abs(t/b)

h=sqrt(h^2-b*h^2.1);
z(i)=a*b;
x(i)=h;
shg
hold on
i=i+1;
a=a+1;

end

plot(z,x)

with $(t=8 , b=0.8 , h=8 a=1),(t=8 , b=0.7 , h=8 a=1),(t=8 , b=0.1 , h=8 a=1)$ to check the result noticed, the relationship

$\surd (h_n)^2-0.8(h_{n})^{2.1}<(h_n)^2-0.7(h_{n})^{2.1}<(h_n)^2-0.1(h_{n})^{2.1}$

Hence consider when b is sufficient small, the derivative exist and bounded all the value with b>dz

$(h+dh)^2-h^2=-h^{2.1}dz$

$2hdh=-h^{2.1}dz$

$\int_{h_0}^{h}-h^{-1.1}=\int_{0}^{z}0.5dz$

obtain

$\frac{1}{h}=\frac{1}{h_0}+0.5z$

when z tend to infinity, h tend to 0 so does all the value of 0.8>b>0 as it is bounded

Answer 1

A graph plot is not a prove about behaviour neear infinity.

Instead, let $y_n=h_n^2$. Then $y_0=64$ and $$ y_{n+1}=y_n-by_n^{1.05}=y_n\cdot(1-by_n^{0.05})$$ looks much simpler. We can prove by induction that $$0<y_{n+1}<y_n $$ so that $\{y_n\}$ is a strictly decreasing sequence bounded from below by $0$ (using crucially that the induction hypothesis and bounds on $b$ imply $0<by_n^{0.05}\le 0.8\cdot 64^{0.05}\approx 0.985$). Hence $y:=\lim_{n\to\infty} y_n$ exists. For this limit, we must clearly have$0\le y<64$ as well as $y=y-by^{1.05})$, i.e., $by^{1.05}=0$ and ultimately $y=0$.