My understanding is that for a sequence to be decreasing $a_{n+1} \leq a_n$. How do I go about using that in regard to the question below? Also to show that it is bounded below by $1$ if that limit is to infinity?
Let $c \geq 1$ and let $a_n = c^{1/n}$ for each $n \geq 1$. Show that the sequence $(a_n)_{n=1}^{\infty}$ is decreasing and bounded below by $1$.
On
To show it's decreasing just compare two arbitrary adjacent terms. I.e $$a_{n+1}-a_n=c^{\frac{1}{n}}(c^{\frac{-1}{n^2+n}}-1) \leq 0$$ using the fact that $c\geq 1$. Now assume that the sequence is not bounded below by $1$. Hence there exists a $\varepsilon>0$ and $n\in \mathbb{N}$ such that $a_n\leq 1-\varepsilon$. Thus $$c^{\frac{1}{n}}\leq 1-\varepsilon$$ $$c\leq( 1-\varepsilon)^n$$ contradicting the fact that $c\geq 1$. So $1$ is indeed a lower bound.
Hint:
To compare two positive numbers, you can compare some relevant power of these numbers.
Yes, the limit is $1$, but it's also bounded from below by $0$, and $0$ is not the limit of $a_n$ as $n$ tends to infinity.