$\lfloor\frac12+\frac1{2^2}+\frac1{2^3}+\cdots\rfloor\;$ vs $\;\lim_{n\to\infty}\lfloor\frac12+\frac1{2^2}+\cdots+\frac1{2^n}\rfloor$ Is there any difference between answers of $[1]$ and $[2]$?
$$\Bigg\lfloor\frac12+\frac1{2^2}+\frac1{2^3}+\cdots\Bigg\rfloor \tag*{$\space.....[1]$}$$
$$
\lim _{n \rightarrow \infty} \Bigg\lfloor\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n}}\Bigg\rfloor \tag*{$ \space.....[2] $}$$
If yes then please do explain that why I can’t write $[2]$ as $[1]$ even if $n$ tends to $\infty$ in $[2]$
(Notice the use of the 'floor' function indicated by the type of brackets.)
NOTE- PLEASE don’t unnecessarily edit $[1]$ and $[2]$. It is exactly  as it should be.
 A: The problem here is that we cannot simply exchange the limit and the function because the floor function is not continuous over $\mathbb R$.
Indeed we know that continuous functions map convergent sequences to convergent sequences. Therefore in that case we have:
$$\lim\limits_{n\to \infty}f(a_n) = f(\lim\limits_{n\to\infty}a_n)$$
As @Jmoravitz remarked this does not have to hold for functions that fail to be continuous, e.g. in our case $f: \mathbb R \to \mathbb N$, defined by $f(x)=\lfloor x \rfloor$. Indeed by direct computation we recognise the geometric series:
$$\left\lfloor\lim_{n \to \infty } \sum_{i=1}^n \frac{1}{2^i}\right\rfloor= \left\lfloor\sum_{i=1}^\infty \frac{1}{2^i} \right\rfloor= \left\lfloor\frac{1}{1-0.5}-1 \right\rfloor=1.$$
However, if we first floor the finite sum, we see that $0<\sum_{i=1}^n \frac{1}{2^i}< 1$ for all $n \in \mathbb N$. This is why we find:
$$\lim_{n \to \infty } \left\lfloor \sum_{i=1}^n \frac{1}{2^i} \right\rfloor= \lim_{n \to \infty } 0=0.$$
We see that we cannot simply exchange the limit and "applying the function".
A: The difference the difference between $\lim\limits_{n\to \infty} f(g(n))$ and $f(\lim\limits_{n\to \infty}g(n))$.
In $\lim _{n \rightarrow \infty} \Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\cdots+\frac{1}{(2)^{n}}\Bigg\rfloor \tag*{$ \space.....[2] $}$ you take a sum, floor it, then take the limits of the floors.
In $\Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\infty\Bigg\rfloor \tag*{$ \space.....[1] $}$  which can is defined as, and can be written as, $\Bigg\lfloor\lim\limits_{n\to \infty}(\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\frac{1}{(2)^{n}})\Bigg\rfloor \tag*{$ \space.....[1] $}$ you take a sum, find its limit and then floor it in the end.
Different things.
.......
$\lim _{n \rightarrow \infty} \Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\cdots+\frac{1}{(2)^{n}}\Bigg\rfloor \tag*{$ \space.....[2] $}=0$
Why?  Because $0< \frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\cdots+\frac{1}{(2)^{n}} < 1 $ for all $n$.  So  $\Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\cdots+\frac{1}{(2)^{n}}\Bigg\rfloor \tag*{$ \space.....[2] $}=0$ for all $n$. So $\lim _{n \rightarrow \infty} \Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\cdots+\frac{1}{(2)^{n}}\Bigg\rfloor \tag*{$ \space.....[2] $}=\lim_{n\to \infty} 0 = 0$.
But $\Bigg\lfloor\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\infty\Bigg\rfloor \tag*{$ \space.....[1] $}$$=\Bigg\lfloor\lim\limits_{n\to \infty}(\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\frac{1}{(2)^{n}})\Bigg\rfloor \tag*{$ \space.....[1] $}=1$
Why?
Because $\lim\limits_{n\to \infty}\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\frac{1}{(2)^{n}}= \lim\limits_{n\to \infty} 1- \frac 1{2^{n}} = 1$.  So $\Bigg\lfloor\lim\limits_{n\to \infty}(\frac{1}{2}+\frac{1}{(2)^{2}}+\frac{1}{(2)^{3}}+\frac{1}{(2)^{4}}+...\frac{1}{(2)^{n}})\Bigg\rfloor \tag*{$ \space.....[1] $}= \Bigg\lfloor 1 \Bigg\rfloor = 1$
