Limit of product of infinite sequences and infinite series If the limit $$\lim_{n\to \infty} n^p u_n =A \tag{1}\\ p\gt 1;A \lt \infty$$ then I have to show that the limit of the series $\sum_{n=0} ^{\infty}u_n $ converges. I know that since $n^p$ blows up at infinity $u_n$ must somehow go to zero. I have tried the following:
$$\lim_{n\to \infty} n^p u_n =A \tag{2} = \lim_{n\to \infty} n^p \cdot \lim_{n\to \infty} u_n \\ 
\therefore$$ $$\lim_{n\to \infty}u_n = \frac{\lim_{n\to \infty}n^p}{A} \tag{3}$$ 
From there I concluded that $u_n = \frac{n^p}{A}$. Then applying the ratio test:
$$\lim_{n\to \infty}\frac{u_{n+1}}{u_n} = \lim_{n\to \infty} {}\frac{(n+1)^p}{n^p} \lt 1 \tag{4}$$ 
but obviously this doesn't work and is wrong. I am not convinced that $(2)$ is correct because I am not sure if the limits ($\lim_{n \to \infty }n^p $ is $\infty$ ???) exist. Also $(4)$ doesn't really work I guess because the limit is probably $=1$.
 A: As $\lim_{n\to \infty} n^p u_n =A $ is finite, $(n^p u_n)$ is a bounded sequence (a sequence that converges is bounded). Let's say
$\vert n^p u_n \vert \le B$ with $B >0$.
then you have $$\vert u_n \vert \le \frac{B}{n^p}$$ proving that $\sum u_n$ is absolutely converging as for $p > 1$, $\sum 1/n^p$ converges.
A: First of all, you are doing a lot of illegal stuff there. (2), as you already pointed out, does not work as $n^p$ is not convergent. Then you are essentially dividing by $0$ from (2) to (3) (among another elementary mistake) and after that, you conclude $u_n = \frac{n^p}{A}$ from $\lim\limits_{n\to \infty}u_n = \frac{\lim\limits_{n\to \infty}n^p}{A}$, which is obviously not correct, either.
Now for your actual question. Write $$\sum_{n=0}^{\infty}u_n = \sum_{n=0}^{\infty} \frac{1}{n^p}(n^pu_n).$$
Can you figure it out now?
A: $(2)$ and $(3)$ are meaningless. $(4)$ is actually right (except that the limit is equal to $1$), but your derivation isn’t. 
So, here, the ratio test is useless. The actual useful assumption is that $n^pu_n$ is convergent. Therefore it is bounded: so $|u_n| \leq Mn^{-p}$ for each $n$. Can you take it from there?
