Jtdylktuy

As the title says, i want to show that $$lim_{nto infty} frac{n^b}{(1+a)^n}=0$$ where $a,bin mathbb{R}$ with $a>0$.

I tried to bound the sequence $x_n = frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^ngeq 1+na$) but then the problem is de numerator $n^b$.

I'm grateful for any hints for doing this problem, or some steps to clarify.

HINT:

Bernoulli's inequality won't suffice when $bge1$, but the binomial theorem will suffice.

From the binomial theorem, with $n>b$, we see that

$$(1+a)^nge binom{n}{lfloor brfloor +1}a^{lfloor b rfloor +1}$$

where $lfloor cdot rfloor$ is the floor function.

Can you proceed?

We need to show that

$$logleft(frac{n^b}{(1+a)^n}right)=blog n-nlog(1+a)=nleft(bfrac{log n}n-log(1+a)right) to -infty$$

that is

$$frac{log n}n to 0$$

By the ratio test, the sequence converges if
$$frac{(n+1)^b}{(1+a)n^b}<1.$$

This is achieved when

$$n>frac1{sqrt[b]{1+a}-1},$$ a finite number.

E.g., with $dfrac{n^3}{2^n}$, for $n>3$,

$$frac{(n+1)^3}{2n^3}<frac{125}{2cdot64}<1.$$