1
$\begingroup$

$$\sum_{i=0}^k 2^{n/{2^i}}$$

I'm trying to find an actual sum of this little nice sum , but I think that there's a problem

with it being a geometric series .

I'd appreciate any help

Regards

  • 0
    Try calculating the $log$ of this expression.2012-04-14
  • 0
    @ofer: The $log$ ? but I do not need the $log$ of this expression , or do I ?2012-04-14
  • 2
    It is not a geometric series. If $x=2^n$, you are looking at $x+\sqrt{x}+\sqrt[4]{x}+\sqrt[8]{x}+\cdots$ (finite sum). No pleasant closed form.2012-04-14
  • 0
    @AndréNicolas: Then what you're saying is that I can't find its final sum ?2012-04-14
  • 1
    @ron: I am at least saying that I cannot! Also, there will not be a closed form in terms of standard functions.2012-04-14
  • 0
    If you can evaluate the value of $log$ of the sum, and you'll get $log Sum = A$, then the original sum is $e^A$.2012-04-15
  • 0
    This is more a guess, but can we approximate by $ \sum_{m=0}^k 2^{n/{2^m}} \sim \int_0^k 2^{n/{2^m}} dm $? Not sure if this makes it easier, but it's different view.2012-07-09

0 Answers 0