Using Comparison test the series for convergence. Stuck with understanding the hint.

Question

I'm currently having some trouble trying to start this question. I'm meant to test the series $\sum_{k=1}^{\infty} \frac{k^3}{3^k}$ for convergence.

The question provided a hint to start off by showing $\lim_{x\to\infty}x^pe^{-ax} = 0$ and use this result to explore the behavior of $\frac{k^3}{(3a)^k}$.

I started off by using L'Hopital's rule by rewriting the limit as $\lim_{x\to\infty} \frac{x^p}{e^{ax}}$ and using L'Hopital's rule I then convert the limit into

$\lim_{x\to\infty} \frac{p(p-1)....(2)(1)}{a^pe^{ax}} = 0$ after applying L'Hopital's rule p times.

for a $p\in N$ and $a>0$. However, I'm not sure how to relate this result to $\frac{k^3}{(3a)^k}$.

I figure that if $0<a<1$ than $ \frac{k^3}{3^k}<\frac{k^3}{(3a)^k}$ and test if the series $\sum_{k=1}^{\infty} \frac{k^3}{(3a)^k}$ converges.

Alternatively if a>1 than $ \frac{k^3}{3^k}>\frac{k^3}{(3a)^k}$ and test if the series $\sum_{k=1}^{\infty} \frac{k^3}{(3a)^k}$ diverges.

However it all comes down to $\frac{k^3}{(3a)^k}$ and I'm not sure what to make of it. I was hoping someone could guide me in understanding how to make use of the hint?

Answer 1

I think the question might be trying to lead you toward an inequality of the form $$\frac{k^3}{3^k} \le C \frac{1}{2^k}.$$ In this case you could establish convergence of your series by a direct comparison with $\displaystyle \sum_{k=1}^\infty \frac{1}{2^k}$.

Since $$k^3 \frac{2^k}{3^k} = k^3 e^{\log (2/3) k}$$ and $\log(2/3) < 0$, you can use the fact that $x^3 e^{\log(2/3)x} \to 0$ to establish that $k^3 \dfrac{2^k}{3^k}$ is bounded above.

Answer 2

The hint means that $e^{ax}$ increases much faster than $x^p$. Hence, show that $\dfrac{k^3}{3^k}$ less than, for example, $\dfrac{1}{k^2}$, when $k$ is large enough.