Geometric series function

Question

The geometric series is defined by :

$$\sum^x_{k=1} ar^k$$

But is there a function that is an extension of the geometric series, that satisfies : $$f(1)=r$$ $$f(x+1)=f(x)+r^{x+1}$$

That works for all $x \in \mathbb{R},\:x > 1$?

Answer 1

By using the partial sums of the geometric series, we have the following identity: $$\sum_{k=1}^n r^k = \sum_{k=0}^n r^k - 1 =\frac{1-r^{n+1}}{1-r} - 1$$

As your function's value is given by $f(n) = \sum_{k=1}^n r^k$ for any positive integer $n$, the formula above is, in some way, the natural extension of this expression to non-integer arguments.

Your function would thus be given by

$$f(x) = \frac{1-r^{x+1}}{1-r} - 1$$

for some fixed $r \neq 1$.

Note that this function inherits its limit from the geometric series that we used to initially define it - if $|r|\in(0,1)$, then $\lim\limits_{x \rightarrow \infty}{f(x)} = \frac{1}{1-r} - 1 = \sum_{k=1}^\infty r^k$. Similarly, it is divergent for $x\rightarrow\infty$ in the case that $|r|>0$, just like a geometric series.

Of course, this is by no means the only solution - in fact, you could multiply it with any function that is periodic with period $1$, like $g(x) = \cos(2\pi x)$, and the result still holds.

Answer 2

Assuming $r \ne 1$, $$ f(x) = g(x) \dfrac{r^{x}-1}{r-1} $$ where $g(x)$ is periodic with period $1$, and $g(1)=r$.