# Need explanation , comparing bounded integrals

#### idontknow

How $$\displaystyle t^n$$ got out of integral as $$\displaystyle n^n$$ ? $$\displaystyle \; n\in \mathbb{N}.$$

$$\displaystyle \int_{n}^{\infty} t^n e^{-t}dt < n^n \int_{n}^{\infty} e^{-t} dt$$.

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#### RDKGames

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.