How is $\displaystyle\sum_{n=1}^{\infty} \frac{n^43^n}{n!}$ absolutely convergent?

Question

For $\displaystyle\sum_{n=1}^{\infty} \frac{n^43^n}{n!}$ we use the ratio test. By $\lim_{n\to\infty}$ my end result is that it goes to $0$ and is convergent.

I do not know how to prove it is absolutely convergent.

Answer 1

What about the ratio test? One has \begin{align*} \limsup_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\rightarrow\infty}\frac{(n+1)^{4}3^{n+1}}{(n+1)!}\times\frac{n!}{n^{4}3^{n}}\\\\ & = \limsup_{n\rightarrow\infty}\frac{3(n+1)^{3}}{n^{4}}\\\ & = \limsup_{n\rightarrow\infty}\frac{3}{n}\times\left(\frac{n+1}{n}\right)^{3}\\\\ & = \limsup_{n\rightarrow\infty}\frac{3}{n}\times\left(1 + \frac{1}{n}\right)^{3} = 0 < 1 \end{align*}

Consequently, the given series converges according to the ratio test.

Hopefully this helps.

Answer 2

The Ratio Test proves absolute convergence. There is no situation where the Ratio Test proves convergence, but not absolute convergence.

Answer 3

As the series is positive it suffices to show that the series itself converges.

$$\sum_{n=0}^{\infty} \frac{n^43^n}{n!} = \\ \sum_{n=0}^{\infty} \frac{(3(n^{\frac{4}{n}}))^n}{n!} \leq \sum_{n=0}^{\infty} \frac{(27)^n}{n!} = e^{27} $$

[Because $n^{\frac{1}{n}}≤\sqrt{3} $] for all $n \in \mathbb N $.

So, we can see that the series is convergent and also absolutely.

Answer 4

Note that for power series if $\sum a_nx^n$ has radius $R$ then so does $\sum n^ka_nx^n$.

Since $e^x=\sum \frac{x^n}{n!}$ has infinite radius, the series converges for $x=3$ and also when multiplying the terms by $n^4$.