How do I show $ 100^{n} $dominates $ n^{2.5} $

Question

So i'm a little shaky on limits and I want to show I could take the limit of$ \frac{x^{2.5}}{100^{x}} $ but it would get very messy with L'Hopitals rule is their an easier way to show that this goes to 0 without having to do a bunch of algebra?

Answer 1

You can show $100^n > n^{5/2}$ for every $n\ge 1$ by induction.

The ratio between successive terms is $100$ for $100^n$ and only $(1+\frac 1n)^{5/2}$ for $n^{5/2}$ so once the inequality holds it can never stop holding.

Answer 2

Remember that

$$\displaystyle\sum_{n} a_{n}< \infty \Rightarrow \lim\limits_{n\rightarrow \infty}a_{n}=0 $$

Observe that $\displaystyle\sum_{n} a_{n}< \infty$, since

$$\displaystyle \lim\limits_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim\limits_{n\rightarrow \infty}\frac{1}{100}(1+\frac{1}{n})^{2.5}<1. $$

Answer 3

Upon taking logarithms from both sides you have $$ 2n> 2.5 \log n$$ or $$\frac {\log n}{n}<0.8$$ Note that $$\lim_{n\to \infty }\frac {\log n}{n} =0$$

Thus the statement is true.