1. ## Find x

Find the value of :
$\lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor$+...........upto $\infty$

2. Without going into a detailed analysis, I am sure that this series diverges for all (fixed) values of x.

3. Hello everyone
Originally Posted by JG89
Without going into a detailed analysis, I am sure that this series diverges for all (fixed) values of x.
For finite positive $x$, all terms after a certain point are zero, so it certainly has a finite sum.

What the answer is, though, I don't know. I'm thinking about it...but I think it may be to do with expressing [an integer] $x$ in binary form.

4. I just clicked on his Latex source and it shows that he`s talking about the floor function for each term. I thought they were square brackets. How embarassing

Hello everyoneFor finite positive $x$, all terms after a certain point are zero, so it certainly has a finite sum.

What the answer is, though, I don't know. I'm thinking about it...but I think it may be to do with expressing [an integer] $x$ in binary form.

The answer is indeed x,if x is a finite positive number.

6. Hello pankaj
Originally Posted by pankaj
The answer is indeed x,if x is a finite positive number.
Sorry, I don't really understand what you mean here.

If $f(x) =\lfloor\frac {x + 1}{2}\rfloor + \lfloor\frac {x + 2}{4}\rfloor + \lfloor\frac {x + 3}{6}\rfloor + \lfloor\frac {x + 4}{8}\rfloor + ...$ then

$f(1) = 1,\, f(2) = 2,\, f(3) = 4,\, f(4) = 5,\, f(5) = 7,\, ...$

and the differences between values of $f(x)$ for consecutive integer values of $x$ don't form any obvious pattern.

However (using a spreadsheet to do the number-crunching) I have found that for all the values of $n$ I've tried so far (which is up to $n = 7$), $f(2^n) = f(2^{n-1}) + 1$. Which is what made me think it might be worth looking at powers of $2$ or the binary representation of $x$.

Further thoughts: as $x$ increases through consecutive integer values:

• the value of the first term, $\lfloor\frac {x + 1}{2}\rfloor$ increases by $1$ every second value of $x$; i.e. when $x =2, 4, 6, 8, ...$
• the value of the second term, $\lfloor\frac {x + 2}{4}\rfloor$ increases by $1$ every 4th value of $x$; i.e. when $x = 2, 6, 10, 14, 18, ...$
• the value of the third term, $\lfloor\frac {x + 3}{6}\rfloor$ increases by $1$ every 6th value of $x$; i.e. when $x = 3, 9, 15, 21, ...$
• ... and so on.