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What would be the value of floor function of $\lim\limits_{N\to\infty}\left\lfloor\sum\limits_{r=1}^N\frac{1}{2^r}\right\rfloor$

would it be $1$ or would it be $0$ ?

The formula I use for this is that of infinite summation series that is $\frac{a}{1-r}$ but I have no clue how to find out what the floor value of the above expression would be.

P.s I am a high school student so please explain in simple terms, and yes I do know basic calculus.

EDIT: I'm sorry it was given $\lim_{N \to \infty}$ in the problem

marks_404
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    $\sum\limits_{r=1}^\infty \frac{1}{2^r} = 1$ and $\lfloor1 \rfloor= 1$. – Asinomás Aug 13 '21 at 15:42
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    To be clear... infinity is not a number and the above expression should be rewritten with limits if you want to be formal. Are you asking for $\left\lfloor \lim\limits_{N\to\infty}\sum\limits_{r=1}^N\frac{1}{2^r}\right\rfloor$ which is equal to $1$? Or are you asking for $\lim\limits_{N\to\infty}\left\lfloor\sum\limits_{r=1}^N\frac{1}{2^r}\right\rfloor$ which is equal to $0$? Note that changing whether the floor is inside or outside of the limit affects the answer. – JMoravitz Aug 13 '21 at 15:44
  • "Greatest integer function" confused me - say "ceil" or "ceiling" to be more specific (and concise). EDIT: It confused me even more since you actually meant the "floor" function! – FShrike Aug 13 '21 at 15:44
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    @FShrike you mean floor. The greatest integer which is less than or equal to an input is otherwise called the "floor" of the number. Ceiling is the other name for the least integer which is greater than or equal to an input – JMoravitz Aug 13 '21 at 15:45
  • @JMoravitz Yes, well spotted. I read "greatest integer" as the first integer greater than... – FShrike Aug 13 '21 at 15:46
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    @JMoravitz limit is outside the "gif brackets". How did u arrive at this answer though? – marks_404 Aug 13 '21 at 15:48
  • @FShrike will change it. – marks_404 Aug 13 '21 at 15:49
  • It's not that important protus since clearly everyone else knew what you meant. I suppose "floor" is just less verbose, and has a natural notation to it: $\lfloor\rfloor$ – FShrike Aug 13 '21 at 15:50
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    "Limit is outside the gif brackets" Then, as alluded to already, every partial sum here is strictly less than $1$ and so floors to zero... just like how $0.9999$ is strictly less than $1$ and so too is $0.\underbrace{9999\dots 9}_{n~\text{times}}$ is strictly less than $1$. You are then investigating the limit as $n\to\infty$ of a constant sequence of zeroes... which is of course zero. – JMoravitz Aug 13 '21 at 15:51
  • There is no such thing as "infinitely smaller than." The expression with the floor brackets on the outside, we leave the floor brackets alone for now and look at the inside. What is on the inside evaluates to identically $1$. Not "almost 1" or "close to 1" but exactly $1$. Once we have evaluated what is inside of the floor brackets we see that the expression simplifies to $\lfloor 1\rfloor$ which is of course $1$, not zero. – JMoravitz Aug 13 '21 at 16:00
  • @JMoravitz got it! – marks_404 Aug 13 '21 at 16:01
  • See also: Is it true that $0.999999\dots = 1$? – JMoravitz Aug 13 '21 at 16:02
  • @JMoravitz yes indeed it is helpful and explains the idea behind the limits being outside or inside the function. – marks_404 Aug 13 '21 at 16:10

2 Answers2

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We know from the geometric series formula that

$$\sum\limits_{i = 1}^N r^i = r \frac{1 - r^{N}}{1 - r}$$

whenever $r \neq 1$. In particular, therefore, we see that

$$\sum\limits_{i = 1}^N \frac{1}{2^i} = \sum\limits_{i = 1}^N (\frac{1}{2})^i = \frac{1}{2}\frac{1 - (1/2)^N}{1 - 1/2} = 1 - (\frac{1}{2})^N$$

Now whenever $N \geq 1$, we see that $0 \leq 1 - (\frac{1}{2})^N < 1$. Therefore, $0 = \lfloor 1 - (\frac{1}{2})^N \rfloor$ for all $N \geq 1$. Therefore, $\lfloor \sum\limits_{i = 1}^N \frac{1}{2^i} \rfloor = 0$ for all $N$.

So the limit is just $\lim\limits_{N \to \infty} 0 = 0$.

Mark Saving
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  • Okay, nice method. Didn't consider this term is there too. So basically if the limits are inside the floor function then $limits_{N \to \infty} (1/2)^N = 0$ and hence [1] = 1$ but if they are outside then it is as you mentioned right? – marks_404 Aug 13 '21 at 16:13
  • If you switch the problem to $\lfloor \lim\limits_{N \to \infty} \sum\limits_{i = 1}^\infty \frac{1}{2^i} \rfloor$, then the answer is indeed 1. Note that it's only possible for us to get a different answer by switching the limit and the floor because the floor function isn't continuous. – Mark Saving Aug 13 '21 at 20:18
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Note that the floor function is not continuous (at integers). So simply taking limit inside is not justified. In fact, in this case, doing so gives you a wrong answer. Here is a sketch:

  1. Use the geometric series formula for finite N to get a closed form expression.
  2. Take the floor. What sequence do you get for every finite N ?
  3. Evaluate the limit. Tell me what answer you got in comments.
Gautam Shenoy
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