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) ?

Consider the the sequence $a_n$ defined as:
If $n=1$ then $a_n=1$, else if $n>1$ then $a_n=-1$.
This is the sequence $a_n = 1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,...$
The Dirichlet inverse (call it $b_n$) of $a_n$ is the oeis sequence called
"Number of ordered factorizations of n" http://oeis.org/A074206 , starting: $b_n = 0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 8,...$, which is also a Dirichlet series. Notice that the first term: $b_0 = 0$.
For reasons I can not explain clearly, $b_n$ has the ordinary generating function:
$$\sum\limits_ {n = 1}^{\infty} b_n r^n = 
 r + \sum\limits_ {a = 
        2}^{\infty} r^{a} + \sum\limits_ {a = 
         2}^{\infty}\sum\limits_ {b = 
        2}^{\infty} r^{ab} + \sum\limits_ {a = 
         2}^{\infty}\sum\limits_ {b = 2}^{\infty}\sum\limits_ {c = 
        2}^{\infty} r^{abc} + \sum\limits_ {a = 
         2}^{\infty}\sum\limits_ {b = 2}^{\infty}\sum\limits_ {c = 
         2}^{\infty}\sum\limits_ {d = 2}^{\infty} r^{abcd} + ... $$
Evaluate each sum partially to the summation index equal to $2$:
$$r + \sum\limits_ {a = 2}^{2} r^{a} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b = 
        2}^{2} r^{ab} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b = 
         2}^{2}\sum\limits_ {c = 
        2}^{2} r^{abc} + \sum\limits_ {a = 2}^{2}\sum\limits_ {b = 
         2}^{2}\sum\limits_ {c = 2}^{2}\sum\limits_ {d = 
        2}^{2} r^{abcd} + ... $$
Simplify:
$$r + r^{2} + r^{2*2} + r^{2*2*2} + r^{2*2*2*2} + r^{2*2*2*2*2} + ...... + r^{2^k}$$
multiply out:
$$r + r^{2} + r^{4} + r^{8} + r^{16} + r^{32} + ...... + r^{2^k}$$
call it $S$ and you are done.
The sum:
$$\sum\limits_ {n = 1}^{\infty} b_n r^n$$
above, probably does not have a closed form other than the multiple sums, and therefore $S$ probably does not have a closed form either. But it points out what the scope of generating functions is.
