100 followers problem

Let $I=\prod_{n\ge 1}\left(1+\frac{1}{\lfloor\sqrt[m]{n}\rfloor}\right)^{(-1)^{n-1}}$ Note that whenever $k^m\le n\le (k+1)^m-1$ , for $k=1,2,\cdots$ , we have $\lfloor\sqrt[m]{n}\rfloor=k$ . Also, note that this happens for $(k+1)^{m}-k^{m}$ number of integers, which is an odd number. Hence, for example, $\prod_{n= k^m}^{(k+1)^m-1}\left(1+\frac{1}{\lfloor\sqrt[m]{n}\rfloor}\right)^{(-1)^{n-1}}=\left(1+\frac{1}{k}\right)^{(-1)^{k-1}}$ Thus, the product becomes $\prod_{k\ge 1}\left(1+\frac{1}{k}\right)^{(-1)^{k-1}}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdots$ which is the famous Walli's product and hence is equal to $\frac{\pi}{2}$ . Thus the answer is $\boxed{2}$ .

$\large \displaystyle \prod _{ n=1 }^{ \infty }{ \left( 1+\frac { 1 }{ \left\lfloor \sqrt [ 100 ]{ n } \right\rfloor } \right) }^{(-1)^{n-1}} =\frac { \pi }{ a }$

Taking logs on both sides of this equation shows us that $\ln{\frac{\pi}{a}} = \displaystyle \sum_{ n=1 }^{ \infty }{ (-1)^{ n-1 } \ln{\left( 1+\frac{1}{ \left\lfloor \sqrt [ 100 ]{ n } \right\rfloor } \right) }}$

Upon expanding this sum, we see that $\displaystyle \ln{\frac{ \pi }{ a }} \space = \space \ln{2} - \ln{\frac{3}{2}} + \ln{\frac{4}{3}} - \cdots$

$\displaystyle \implies \ln{\frac{\pi}{a}} \ = \sum_{n=1}^{\infty}{ (-1)^{n-1} \ln{\frac{n+1}{n}} } \ = \ \ln{ \left( 2 \times \frac{2}{3} \times \frac{4}{3} \times \cdots \right) }$

The product contained in the logarithm above is the Wallis' product, which evaluates to $\dfrac{\pi}{2}$ .

Therefore, $\ln{\dfrac{\pi}{a}} \ = \ \ln{\dfrac{\pi}{2}}$ so $a \ = \ 2$ .