Generalizations of $\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}$ and $\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}$ I noticed that the following identies hold with the help of wolfram alpha and oeis. I'm sure they're well-known, but I'd like to know how they generalize.
$$\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}=(-1)^n(a_n-b_n\sqrt5),$$
where $\phi=\dfrac{\sqrt5-1}2,a_0=1,b_0=0,$ and $a_n$ and $b_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt5=2+\dfrac1{4+\dfrac1{4+\dfrac1{4+\cdots}}}$
Similarly,
$$\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}=(-1)^n\dfrac{c_{13n}-d_{13n}\sqrt2}{\sqrt2},$$
where $c_0=1,d_0=0,$ and $c_n$ and $d_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt2=1+\dfrac1{2+\dfrac1{2+\dfrac1{2+\cdots}}}$
Do similar identities exist for other values of $\dfrac{\sqrt{k^2+4}-k}2$ for positive integer $k$ (the previous examples are the cases $k=1,2$ respectively)?
 A: This is a really awesome observation of yours. The main thing that will help you here is to think in purely algebraic terms. Notably, any product, sum, quotient, etc. involving only rational numbers and $\sqrt{k}$ can be written in the form $a+b\sqrt{k}$. In particular, if we define $\phi_k=\frac{\sqrt{k^2+4}-k}2$, notice that this is a root of
$$\phi_k^2+k\phi_k-1=0.$$
Next, define the function
$$f_k(n)=\sum_{m=n}^{\infty}\phi_k^m=\frac{\phi_k^m}{1-\phi_k}$$
where the latter expression is a well-known identity for geometric sums. We want to find two rational sequence $a_n$ and $b_n$ such that
$$f_k(n)=a_n+b_n\phi_k$$ 
This would be a very long post if we went through all the algebra to derive all the equations I'm gonna use (I started to write such a post) - so I won't where the calculation is routine. From the quadratic equation above, it follows that
$$f_k(0)=\frac{1}{1-\phi_k}=\frac{k+1}k+\frac{1}k\phi_k$$
which can be taken to mean that $a_0=\frac{k+1}{k}$ and $b_0=\frac{1}k$. Further, noting that $f_k(n+1)=\phi_kf_k(n)$, we write
$$a_{n+1}+b_{n+1}\phi_k=a_n\phi_k + b_n\phi_k^2$$
and, after rearranging terms and seeing that $\phi_k^2=1-k\phi_k$, we get
$$a_{n+1}+b_{n+1}\phi_k=b_n+(a_n-kb_n)\phi_k$$
implying that $a$ and $b$ are governed by the following recurrence
$$\begin{array}{ll}a_0=\frac{k+1}{k} && b_0=\frac{1}k \\ a_{n+1}=b_n && b_{n+1}=-kb_n+a_n \end{array}$$
wherefrom we can easily eliminate $a$ to get the equations
$$b_0=\frac{1}k$$
$$b_1=\frac{1}k$$
$$b_{n+2}'=-kb_{n+1}'+b_n'$$
And then we have $f_k(n)=b_{n-1}+b_n\phi_k$, which is a pretty interesting relation in and of itself, especially given how $b$ looks kind of like the Fibonacci sequence. Actually, if $k=1$, then, after $3$ places, the sequence $b$ is the Fibonacci sequence, just with its signs alternating. However, to further elucidate the connection, set $b_n'=k(-1)^nb_n$. Then we have
$$b_0'=1$$
$$b_1'=-1$$
$$b_{n+2}'=kb_{n+1}'+b_n'$$
which, for $k=1$ is a shift of the Fibonacci sequence. Also, we have $f_k(n)=(-1)^n(b_n'\phi_k-b_{n-1}')$
Better yet, this series very directly relates to the series of convergents for $\phi_k$. Another property of $\phi_k$ following from the quadratic it solves is that
$$\phi_k=\frac{1}{k+\phi_k}$$
meaning that the continued fraction for $\phi_k$ is $[0;k,k,\ldots]$ and, therefore, that the series of convergents $x$ can be defined by
$$x_0=0$$
$$x_{n+1}=\frac{1}{k+x_n}$$
and that, if we substitute $x_n=\frac{p_n}{q_n}$ for coprime sequences of integers $p$ and $q$, we get
$$\frac{p_n}{q_n}=\frac{0}{1}$$
$$\frac{p_{n+1}}{q_{n+1}}=\frac{1}{k+\frac{p_n}{q_n}}=\frac{q_n}{kq_n+p_n}$$
which, again, yields a nice recurrence relation
$$\begin{array}{ll}p_0=0 &&q_0=1\\p_{n+1}=q_n && q_{n+1}=kq_n+p_n\end{array}$$
which looks fairly familiar, especially when we eliminate $p$ from the equation and get
$$q_0=1$$
$$q_1=k$$
$$q_{n+2}=kq_{n+1}+q_n$$
where the convergents are just $x_n=\frac{q_{n-1}}{q_n}$ - but, hey, isn't that the same recurrence relation as defines $b'$, just with different boundary conditions? (hint: yes)
From here, note that, for $k=1$, then $q_n=(-1)^{n+1} b_{n+3}$ and $p_n=(-1)^n a_{n+3}$, so that means that $f_1(n+3)=(-1)^n(p_n-q_n\phi_1)$ where $p_n/q_n$ are the convergents of $\phi_1$. Presumably, the exact relation you cite, using $\sqrt{5}$ is just a stone's throw from there. Unfortunately, far as I can tell, there is no shift of $b$ that would align it with $q$ for any $k>1$, but perhaps using this groundwork, where you have two similar recurrence relations you can find something. (If I think of something, I'll update this answer; I suspect that, given the relation of $a_n$ and $b_n$, we could easily find $c_n$ and $d_n$ such that $a_n+b_n\phi_k=c_n+d_n\phi_k'$ for some other irrational $\phi_k'$ satisfying certain properties and there might be some trick therein to derive an identity; I would tend to believe that something more exists)
A: Well, I came up with a much better solution than my previous one, but it's too substantial to add to my old answer.
We are looking for irrational numbers $x$ with convergents $\frac{p_n}{q_n}$ such that, if we define $f_x(n)=p_n-q_nx$, there exists some real number $c$ such that $f_x(n+1)=cf_x(n)$. Notice that, according to Wikipedia, if $x$ has continued fraction $[a_0;a_1,a_2,\ldots]$, then the following recurrences hold
$$p_n=a_np_{n-1}+p_{n-2}$$
$$q_n=a_nq_{n-1}+q_{n-2}$$
So, using these, we have
$$f_x(n+2)=a_{n+2}f_x(n+1)+f_x(n)$$
$$c^2f_x(n)=ca_{n+2}f_x(n)+f_x(n)$$
Or, since $f_x(n)$ can never be $0$ since $x$ is irrational, we divide out $f_x(n)$ and receive
$$c^2-ca_{n+2}-1=0$$
$$\frac{c^2-1}{c}=a_{n+2}$$
which must hold for any $n\geq 0$, so the continued fraction of $x$ must be of the form $[y;z,k,k,\ldots]$ for integers $x$, $y$, and $z$. In particular, since the continued fraction of $\phi_k=\frac{\sqrt{k^2+4}-k}2$ is $[0;k,k,\ldots]$ since is the unique value in $[0,1)$ satisfying $\phi_k=\frac{1}{k+\phi_k}$, this means that any real $x$ such that the "residue" of the convergents $f_x(n)$ is an exponential function must be of the form
$$x=y+\frac{1}{z+\phi_k}.$$
In fact, every such $x$ satisfies the condition; to show that, note that, here $f_x(0)=\frac{1}{z+\phi_k}$ and $f_x(1)=1-\frac{z}{z+\phi_k}$ and that the condition $cf_x(0)=f_x(1)$ holds whenever $c=-\phi_k$, which is obvious, given that $c$ satisfies $c^2-kc-1=0$ and $\phi_k$ satisfies $\phi_k^2+k\phi_k-1=0$ and both are the unique roots of their respective polynomials in $(-1,1)$. Thus, inductively, we have that $x=y+\frac{1}{z+\phi_k}$ is a necessary and sufficient condition for $f_x(n+1)=cf_x(n)$ for all $n$, and moreover, this implies
$$f_x(n)=(-\phi_k)^nf(0)=(-\phi_k)^n\frac{1}{z+\phi_k}.$$
Knowing this, and the fact that a geometric sum is also a exponential function, your identities may be proven with ease and further similar ones can be found.
I found an infinite family of solutions using Mathematica to look for solutions to the following equation $f(0)=\frac{1}{z+\phi_k}=\frac{\phi_k^a}{1-\phi_k}=\sum_{m=a}^{\infty}\phi_k^m$ for integer $a$ and $z$. There could be identities not found by solving this equation, but from the solution this has for $a=1$ and $z=k-1$ we get the following family: Let $X_k=\frac{1}{k-1+\phi_k}$ (or that, plus any integer; turns out the integer term doesn't matter) and $p_n/q_n$ be the convergents thereof in lowest form. Then
$$\sum_{m=n+1}^{\infty}\phi_k^m=(-1)^{n+1}(p_n-q_nX_k).$$
I also found ($a=2$,$z=3$) that for $X=\frac{1}{3+\phi_2}=1-\frac{1}{\sqrt{2}}$ with convergents $p_n/q_n$, we have
$$\sum_{m=n+2}\phi_2^m=(-1)^{n+1}(p_n-q_nX).$$
I've not found any other solutions, aside from the ones you list.
A: I got $\displaystyle \sum_{3n+2}^\infty \phi^m = \frac{\phi^{3n+2}}{1- \phi}$ and then I could not find a good way to simplify $1 - \phi$:
$$ 1 - \phi = \frac{3-\sqrt{5}}{2}$$
To get you started, use the identity $\phi^2 = -\phi + 1$, since you have written $\phi = \tfrac{\sqrt{5}-1}{2}$.  And say:
$$ \frac{1}{1-\phi} = \frac{1+\phi}{1 - \phi^2}  = \frac{1+\phi}{\phi}= 1 + \frac{1}{\phi}$$
Zeckendorf's theorem talks about the expansion of number in the Fibonacci Base.  I am using this to try to get the best simplification.
So the sum of your geometric series is $\phi^{3n+2}(1 + \phi^{-1})$.
