$lfloorfrac12+frac1{2^2}+frac1{2^3}+cdotsrfloor;$ vs $;lim_{ntoinfty}lfloorfrac12+frac1{2^2}+cdots+frac1{2^n}rfloor$

The difference the difference between $limlimits_{nto infty} f(g(n))$ and $f(limlimits_{nto infty}g(n))$.

In $lim _{n rightarrow infty} Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+cdots+frac{1}{(2)^{n}}Biggrfloor tag*{$ space.....[2] $}$ you take a sum, floor it, then take the limits of the floors.

In $Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...inftyBiggrfloor tag*{$ space.....[1] $}$ which can is defined as, and can be written as, $Bigglfloorlimlimits_{nto infty}(frac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...frac{1}{(2)^{n}})Biggrfloor tag*{$ space.....[1] $}$ you take a sum, find its limit and then floor it in the end.

Different things.

.......

$lim _{n rightarrow infty} Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+cdots+frac{1}{(2)^{n}}Biggrfloor tag*{$ 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 $Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+cdots+frac{1}{(2)^{n}}Biggrfloor tag*{$ space.....[2] $}=0$ for all $n$. So $lim _{n rightarrow infty} Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+cdots+frac{1}{(2)^{n}}Biggrfloor tag*{$ space.....[2] $}=lim_{nto infty} 0 = 0$.

But $Bigglfloorfrac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...inftyBiggrfloor tag*{$ space.....[1] $}$$=Bigglfloorlimlimits_{nto infty}(frac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...frac{1}{(2)^{n}})Biggrfloor tag*{$ space.....[1] $}=1$

Why?

Because $limlimits_{nto infty}frac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...frac{1}{(2)^{n}}= limlimits_{nto infty} 1- frac 1{2^{n}} = 1$. So $Bigglfloorlimlimits_{nto infty}(frac{1}{2}+frac{1}{(2)^{2}}+frac{1}{(2)^{3}}+frac{1}{(2)^{4}}+...frac{1}{(2)^{n}})Biggrfloor tag*{$ space.....[1] $}= Bigglfloor 1 Biggrfloor = 1$

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: $$limlimits_{nto infty}f(a_n) = f(limlimits_{ntoinfty}a_n)$$ 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: $$leftlfloorlim_{n to infty } sum_{i=1}^n frac{1}{2^i}rightrfloor= leftlfloorsum_{i=1}^infty frac{1}{2^i} rightrfloor= leftlfloorfrac{1}{1-0.5}-1 rightrfloor=1.$$ 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: $$lim_{n to infty } leftlfloor sum_{i=1}^n frac{1}{2^i} rightrfloor= lim_{n to infty } 0=0.$$ We see that we cannot simply exchange the limit and "applying the function".