$\displaystyle \sum_{n=1}^{\log_2x}$...
When you have a sum like this, what is the result of $\log2_x$ since it may return a decimal/fractional value? Is it truncated, ceiled or floored?
I know I could do:
$\displaystyle \sum_{n=1}^{\lceil\log_2x\rceil}$ or $\displaystyle \sum_{n=1}^{\lfloor\log_2x\rfloor}$
to make it a whole number(integer) without any fuzz, but I've seen similar notation before without truncation or ceiling. So I havent yet found out what what the value goes to (when this summation stops).
The usual convention is that $$ \sum_{n=1}^{N} a_n $$ is the sum of all $a_n$ with indices between the lower and upper limits of summation. Hence $$ \sum_{n=1}^{N} a_n = \sum_{1\le n \le N} a_n = \sum_{n=1}^{\lfloor N \rfloor} a_n = a_1 + a_2 + \dotsb + a_{\lfloor N \rfloor}. $$ That being said, if you are interested in what happens as $N$ goes to infinity (as is often the case), it doesn't really matter which convention you choose.
The following is probably overkill, but the above can actually be justified in a meaningful way via a branch of mathematics called measure theory: a series is a special kind of integral. Indeed, if you know any measure theory, a series is an integral of the form $$ \sum_{n=1}^{N} a_n = \int_{[0,N]} a(x) \,\mathrm{d}\mu(x), $$ where $a : \mathbb{R} \to \mathbb{R}$ (or some other codomain) such that $$ a(x) = \begin{cases} a_n & \text{if $x = n \in \mathbb{N}$, and} \\ 0 & \text{otherwise}, \end{cases} $$ and $\mu$ is counting measure on $\mathbb{N}$. If $N$ is not an integer, then \begin{align} \sum_{n=1}^{N} a_n &= \int_{[0,N]} a(x)\,\mathrm{d}\mu(x) && \text{(by definition)} \\ &= \int_{[0,\lfloor N \rfloor]} a(x) \, \mathrm{d}\mu(x) + \int_{(\lfloor N\rfloor, N]} a(x) \,\mathrm{d}\mu(x) && \text{(additivity of integrals)} \\ &= \int_{[0,\lfloor N \rfloor]} a(x) \, \mathrm{d}\mu(x) \tag{1}\\ &= \sum_{n=1}^{\lfloor N \rfloor} a_n. && \text{(definition)} \end{align} The only mystery here is at (1). Here, we are using the fact that the measure of an interval which contains no integers must be zero. That is, $$ \mu((\lfloor N \rfloor, N]) = 0. $$ In a measure space, integration over a set of measure zero is always zero. Hence $$ \int_{(\lfloor N\rfloor, N]} a(x) \,\mathrm{d}\mu(x) = 0. $$ In short, the identity above is justified, that is $$ \sum_{n=1}^{N} a_n = \sum_{n=1}^{\lfloor N \rfloor} a_n. $$