Is $\sum_{n=1}^{\infty}\frac{2^nn!}{(n+1)!}$ absolutely convergent?

Question

I'm very uncomfortable with factorials just because I haven't done many of them. But my basic understanding is if I start with (for example) $(n+1)!$ then this is equivalent to $(n+1)*(n)$ and if it were $(n-1)!$ then this is equivalent to $(n-1)*(n-2)*(n-3)...$

Is my understanding correct? If so did I solve this correctly using the ratio test?

$$\sum_{n=1}^{\infty}\frac{2^nn!}{(n+1)!}$$

So I did the following:

$$|\frac{2^{n+1}(n+1)!}{(n+2)!}*\frac{2^nn!}{(n+1)!}|$$

When I simplified this I got:

$$\frac{2(n+1)!}{(n+2)!}$$

And further...

$$\frac{2}{(n+2)!}$$

And then I took the limit of this and found it to be $0$

Is this correct?

Answer 1

Note that, since limit of the summand

$$ \lim_{n\to \infty} \frac{2^n n!}{(n+1)!}=\infty, $$

then the series diverges. See a related problem.

Answer 2

$$ \frac{n!}{(n+1)!} = \frac1{n+1} .$$ Also, to do the ratio test, you look at the limit of $|a_{n+1}/a_n|$ - you took the product instead.

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{% \sum_{n = 1}^{N}{2^{n}n! \over \pars{n + 1}!} = {1 \over 2}\sum_{n = 2}^{N + 1}{2^{n} \over n}}.\quad$ However, when $x > 0$, $\ds{% {2^{x} - 2^{0} \over x - 0} = 2^{\xi}\ln\pars{2}\,,\qquad 0 < \xi < x}$ $$ \ln\pars{2} < {2^{x} - 1 \over x} < 2^{x}\ln\pars{2} \quad\imp\quad {2^{n} \over n} > {1 \over n} + \ln\pars{2} $$