Power series of $\frac{x+1}{x^2+5}$

I have this problem: write down the power series of $$\frac{x+1}{x^2+5}$$ and specify for which interval the equality (function = series) holds true.

I noticed that I can use differentiation method here.

So I started by writing down:

$$\frac{1}{\sqrt{5}}\arctan \frac{x}{\sqrt{5}} = \sum_{n=0}^\infty \frac{1}{5}\left(\frac{-1}{5}\right)^n\frac{x^{2n+1}}{2n+1} \tag{1}$$

This is known since the power series of $\arctan{u}$ is known.
This equality holds true for $|x| \le \sqrt{5}$ (this I know from theory).

Differentiating (1) gives me:

$$\frac{1}{x^2+5} = \sum_{n=0}^\infty \frac{1}{5}\left(\frac{-1}{5}\right)^n x^{2n}\tag{2}$$

Multiplying this by $x$ gives me:

$$\frac{x}{x^2+5} = \sum_{n=0}^\infty \frac{1}{5}\left(\frac{-1}{5}\right)^n x^{2n+1}\tag{3}$$

Now I need to sum up (2) and (3). I get this somewhat strangely written series:

$$\frac{x+1}{x^2+5} = \sum_{n=0}^\infty \frac{1}{5}\left(\frac{-1}{5}\right)^{\lfloor{n/2}\rfloor} x^n \tag{4}$$

I say strangely written because I use $\lfloor{n/2}\rfloor$ here.

OK, now I know (4) holds true for sure in $(-\sqrt{5}, \sqrt{5})$ (since (2) and (3) have the same radius of convergence as (1)).

Finally, I think that for $x = \pm\sqrt{5}$ equality (4) does not hold true, because if we look at the terms with e.g. odd indices in (4) they are $1/5, -1/5, 1/5, -1/5, ...$ So the sub-sequence of these terms doesn't go to zero. So there's no way (4) can be convergent for $x = \pm\sqrt{5}$. Right?

I hope I didn't miscalculate somewhere but I think I didn't.

My questions are:

1. Is (4) correct? And is there a simpler way to write down (4) without $\lfloor{n/2}\rfloor$? WolframAlpha doesn't help me much for this:

WA power series

I mean, it seems it doesn't provide clear expression for $a_n$ as I do in (4).

2. Is my conclusion for $x=\pm\sqrt{5}$ correct? I mean, that the series (4) diverges there.

3. And also in general... did I make any mistake at any step here?

From the geometric series,$$\frac1{x^2+5}=\frac15\sum_{k=0}^\infty\left(-\frac{x^2}5\right)^k$$ so that

$$\frac{x+1}{x^2+5}=\frac15\sum_{k=0}^\infty(-1)^k\frac{x^{2k}+x^{2k+1}}{5^k}.$$

You can indeed replace the even/odd exponents $2k$ and $2k+1$ by a single sum on the integer $n$, and that makes $k=\lfloor\frac n2\rfloor$, generating the duplicated sequence $0,0,1,1,2,2,\cdots$