Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if $k$ is given) ?

Although I cannot actually give an answer, I have explored many of the avenues for approximation of the sum, which I have illustrated below.
Integral Approximation of the Sum
$$ \int_{0}^{\infty} r^{2^x} \mathrm dx \le \sum_{i=0}^\infty r^{2^i}\le \int_{-1}^{\infty} r^{2^x} \mathrm dx $$
We have:
$$ \int r^{2^x} \mathrm dx = \frac{\text{Ei}\left(2^x \log (r)\right)}{\log (2)} $$
For $r < 1$, this converges to $0$ as $x\to\infty$. Therefore, we have the following bounds:
$$ -\frac{\text{Ei}\left(\log (r)\right)}{\log (2)} \le \sum_{i=0}^\infty r^{2^i} \le -\frac{\text{Ei}\left(\frac{\log (r)}{2}\right)}{\log (2)}  $$
As an example, consider $r = 1/2$. Evaluating the bounds, we have:
$ 0.546307 \le 0.816422 \le 1.15583 $. It is clear that this bound may not be entirely the best. However, as $r\to 1$, the bounds are asymptotically equal.
If we do not let $x \to \infty$ and instead consider the general case, we have the following bounds (if we stop at $c$):
$$ \frac{\text{Ei}\left(2^{c+1} \log (r)\right)-\text{Ei}(\log (r))}{\log (2)} \le \Sigma \le \frac{\text{Ei}\left(2^c \log (r)\right)-\text{Ei}\left(\frac{\log (r)}{2}\right)}{\log (2)} $$
Here is a graph of these bounds.

Error Approximation of the Sum
For the partial sum, we have:
$$ (n-m) r^{2^n} \le \sum _{k=m+1}^n r^{2^k} \le \frac{r^{2^{m+1} (n-m+1)}-r^{2^{m+1}}}{r^{2^{m+1}}-1} $$
For $m = -1$, we have:
$$ (n + 1) r^{2^n} \le \sum_{k=0}^n r^{2^k} \le \frac{r^{n+2}-r}{r-1} $$
The partial sum is necessary if you want an approximation anything near the real thing.
We can use this partial sum to truncate the error of the function. For example, let's say we want to find:
$$ \sum_{k=0}^{n} r^{2^k} $$
With an error no more than $\epsilon$. We split this into a partial sum:
$$ \sum_{k=0}^{m} r^{2^k} + \sum_{k=m+1}^{n} r^{2^k} = \sum_{k=0}^{n} r^{2^k} $$
We want to find $m$ such that the partial sum is less than $\epsilon$.
$$ \sum_{k=m+1}^{n} r^{2^k} \le \epsilon $$
Apply our bounds to get:
$$ \frac{r^{2^{m+1} (n-m+1)}-r^{2^{m+1}}}{r^{2^{m+1}}-1} \le \epsilon $$
Which, given $n, r,\text{and}, \epsilon$ reduces our problem to a 1D search.
