Determine if the series is abs. convergent, conditionally convergent, or divergent

Question

The series is $$\sum_{n=1}^\infty \frac{(-1)^n(2n+3)}{n!}$$ First time I solved this I took the absolute value and then split the series at the "$+$" sign and made it into two parts, solving them separately like so $$2\sum_{n=1}^\infty\frac{n}{n!}+3\sum_{n=1}^\infty\frac{1}{n!}$$ My final answer after taking the limits respectfully was absolute convergence but, the method my professor wants me to use is the ratio test. So my question is what would be the first few steps in solving it like that?

P.S.- I am asking a new question on an old post of mine because I ran out of new question to ask in a 24 hour period.

Answer 1

Yes, but it looks a bit strange, since $(x-4)^{n-1}$ is not a monomial if $n=0$. Of course, this is not a really serious problem, since it is multiplied by $N$ and therefore it vanishes, but it is more natural to write the answer as$$f'(x)=\sum_{n=1}^\infty(-1)^nn\frac{(x-4)^{n-1}}{n+1}$$or as$$f'(x)=\sum_{n=0}^\infty(-1)^{n+1}(n+1)\frac{(x-1)^n}{n+2}.$$

Answer 2

Yes it is correct for $|x-4|<1$

Answer 3

Looks alright to me except the derivative starts at $n=1$ (at least it’s safer so you don’t have to deal with weird expressions if for example $x=4$) and don’t forget to show where this series converges (uniformly; but you get this automatically for every point in the open disk around $x=4$ with the radius of convergence). Otherwise you can‘t pull the $\frac{d}{dx}$ operator into the series (allowing you to differentiate each term of the series on its own)

Let $c_n := \frac{(-1)^n}{n+1}$ then $$f(x) = \sum\limits_{n=0}^\infty c_n (x-4)^n$$

This converges for all $x\in B_1(4)$ by the ration test. Therefore it‘s differentiable in this set with

$$f‘(x) = \sum\limits_{n=1}^\infty c_n n (x-4)^{n-1}$$