Pretty Nice Limit

First we solve the series :- $\displaystyle \sum_{r=3}^{\infty}\frac{(-1)^{r-1}}{r} = \ln(2) - \frac{1}{2}$ This can be done using the taylor series expansion for $\large \ln(1+x)$

$\displaystyle \ln(1+x) = \sum_{r=1}^{\infty}\frac{(-1)^{r-1}}{r}$

So we have : - $\displaystyle \lim_{n\to\infty} \left(n(\sqrt[n]{4} -1) - \ln(4) + 1 \right)^{2n}$

Now it is easy to see that it is $1^{\infty}$ form . As $\displaystyle \lim_{n\to\infty}(n(\sqrt[n]{4} -1) = \lim_{n\to\infty}\frac{(\sqrt[n]{4} -1)}{\frac{1}{n}} = \lim_{x\to 0^{+}}\frac{(4^{x}-1)}{x} = \ln(4)$

So we have $\huge e^{\lim_{n\to\infty} \left(n(\sqrt[n]{4} -1) - \ln(4)\right){2n}}$

Now making a substitution of $\displaystyle x = \frac{1}{n}$ we have :-

$\huge e^{\lim_{x\to 0^{+}}\left(\frac{4^{x}-1}{x} -\ln(4)\right)\frac{2}{x}}$

Now we know $\large 4 = e^{\ln(4)}$ So $\large 4^{x} = e^{x\ln(4)}$

Now we know the taylor series expansion for $\large e^{z} = \sum_{r=0}^{\infty} \frac{z^{r}}{r!}$

We have $\huge e^{\lim_{x\to 0^{+}} \left(\frac{(1+\frac{x\ln(4)}{1!} + \frac{(x\ln(4))^{2}}{2!} + O((x\ln(4))^{3}))-1}{x} - \ln(4)\right)\frac{2}{x}}$

$\huge = e^{\lim_{x\to 0^{+}} \left(\frac{(\frac{x\ln(4)}{1!} + \frac{(x\ln(4))^{2}}{2!} + O((x\ln(4))^{3}))}{x} - \ln(4)\right) \frac{2}{x}}$

$\huge = e^{\lim_{x\to 0^{+}} (\frac{x(\ln(4))^{2}}{2!})\frac{2}{x}}$

$\huge = e^{(\ln(4))^{2}}$

$\large = 4^{\ln(4)}$

Here $O(\cdot)$ denotes the big O notation

$\begin{aligned} L & = \lim_{n \to \infty} \left(n \left(\sqrt[n]4-1\right) - 2\sum_{r=3}^\infty \frac {(-1)^{r-1}}r \right)^{2n} \\ & = \lim_{n \to \infty} \left(\blue{\frac {\sqrt[n]4-1}{\frac 1n}} - 2 \left(\sum_{r=1}^\infty \frac {(-1)^{r-1}}r -1 + \frac 12 \right)\right)^{2n} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies}} \\ & = \lim_{n \to \infty} \left(\blue{\ln 4} - \ln 4 + 1 \right)^{2n} & \small \blue{\text{Differentiate up and down w.r.t. }n} \end{aligned}$

This means that $L$ is of the form $1^\infty$ , and we can use the $1^\infty$ -limit method and we can write $\displaystyle \lim_{x \to \infty} f(x)^{g(x)} = e^{\lim_{x \to \infty} g(x)(f(x) - 1)}$ .

$\begin{aligned} L & = \exp \left(\lim_{n \to \infty} 2n\left(n\left(\sqrt[n]4-1\right) - \ln 4 \right) \right) & \small \blue{\text{where }\exp(x) = e^x} \\ & = \exp \left(\lim_{n \to \infty} \frac {2n\left(\sqrt[n]4-1\right) - 2 \ln 4}{\frac 1n} \right) & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies}} \\ & = \exp \left(\lim_{n \to \infty} \frac {2\left(4^\frac 1n-1\right)-\frac 2n \ln 4 \cdot 4^\frac 1n}{- \frac 1{n^2}}\right) & \small \blue{\text{A 0/0 case again after differentiation}} \\ & = \exp \left(\lim_{n \to \infty} \frac {-\frac 2{n^2} \ln 4 \cdot 4^\frac 1n + \frac 2{n^2} \ln 4 \cdot 4^\frac 1n + \frac 2{n^3} \ln^2 4 \cdot 4^\frac 1n}{\frac 2{n^3}}\right) & \small \blue{\text{Differentiate up and down w.r.t. }n} \\ & = \exp \left(\lim_{n \to \infty} \ln^2 4 \cdot 4^\frac 1n \right) = e^{\ln^2 4} \end{aligned}$

Therefore $\lfloor 100A \rfloor = \lfloor 100 e^{\ln^2 4} \rfloor = \boxed{683}$ .

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Reference: L'Hôpital's rule