1. ## Find x

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

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

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

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

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

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

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

7. Originally Posted by pankaj
The answer is indeed x,if x is a finite positive number.
I am wrong.The post may please be ignored.

8. Is the answer equal to infinite?