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Consider the following theorem:

Let $p \in \mathbb{Z}$ and $\alpha \in (0,1)$. Then $n^p \alpha^n \rightarrow 0$.

To prove this, we will first note that $$ r_n := \frac{a_{n+1}}{a_n} = \frac{(n+1)^p\alpha^{n+1}}{n^p\alpha^n} = \left( 1 + \frac 1 n \right)^p \alpha \rightarrow \alpha < 1 $$ by the Algebra of Limits. Thus far, I understand this proof without problem. However, it now goes on to say that, from the above, we can tell that the sequence is strictly decreasing. This is the part of the proof that I do not understand. How does the above tell us that the sequence is decreasing?

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    Consider two positive numbers, $a$ and $b$. Then $a < b $ iff $a / b < 1$. – 2017-01-27
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    basically you wrote $\frac{a_{n+1}}{a_n}<1$. It's easy to see that $a_n>0$ for each $n$, so multiply both sides by $a_n$ to get the result – 2017-01-27

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It's not true that the sequence is strictly decreasing. What is true is that it is eventually strictly decreasing. That is, there exist $N$ such that $a_N>a_{N+1}>a_{N+2}>\cdots.$

To prove this, we need to use the definition of what it means for $r_n$ to converge to $\alpha$. This means that for any $\epsilon>0$, there exists $N$ such that $|r_n-\alpha|<\epsilon$ for all $n\geq N$. In particular, let's pick $\epsilon=1-\alpha$. Then we get an $N$ such that $|r_n-\alpha|<1-\alpha$ for $n\geq N$, which implies $r_n<1$. (A shorter intuitive version of this argument is: since $r_n$ is getting close to $\alpha$ and $\alpha<1$, $r_n$ must eventually always be less than $1$.)

So we have $r_n<1$ for all $n\geq N$. Since $a_{n+1}=r_na_n$ (and $a_n>0$), this means $a_{n+1}

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Note that it's only true that the sequence is strictly decreasing for sufficiently large $n$; take for instance $p=100, \alpha=\frac12$.

To prove your statement with the 'for sufficiently large $n$' qualifier, consider the following: since your sequence of ratios $r_n$ converges to $\alpha\lt 1$, you can pick some $\beta$ with $\alpha\lt\beta\lt 1$ and show that there's some $N_0$ with $r_n\lt\beta$ for all $n\gt N_0$ (why?). Now $a_{n+1}=r_na_n\lt\beta a_n\lt a_n$ for those $n$.