Need explanation , comparing bounded integrals

Dec 2015
972
128
Earth
How \(\displaystyle t^n \) got out of integral as \(\displaystyle n^n\) ? \(\displaystyle \; n\in \mathbb{N}.\)

\(\displaystyle \int_{n}^{\infty} t^n e^{-t}dt < n^n \int_{n}^{\infty} e^{-t} dt \).
 
Last edited:
Jun 2016
11
14
UK
Not sure you have the right inequality there, surely it's $>$ ?

For all $n \in \mathbb{N}$ and $n < t < \infty$ we have that $t^n > n^n$ which is what is being used here.