A calculus problem by Kenny Lau

Actually, $a_n=(-1)^n\dbinom{-0.5}{n}$ .

For integer-parametered binomial coefficients, we have $\dbinom{n}{r}=\dbinom{n}{r-1}\times\dfrac{n-r+1}{r}$ .

This can be generalized to binomial with non-integer parameters.

If $n=-0.5$ , then $\dfrac{n-r+1}{r}$ becomes $\dfrac{-0.5-r+1}{r}=\dfrac{0.5-r}{r}=(-1)\dfrac{2r-1}{2r}$ .

That means, $\dbinom{-0.5}{n}=\dbinom{-0.5}{n-1}\times(-1)\dfrac{2r-1}{2r}$ .

Therefore, $a_n=(-1)^n\dbinom{-0.5}{n}$ .

Therefore:

$\displaystyle\quad\sum_{n=0}^\infty a_nb_n$

$\displaystyle=\sum_{n=0}^\infty (-1)^n\dbinom{-0.5}n\frac1{2n+1}$

$\displaystyle=\sum_{n=0}^\infty (-1)^n\dbinom{-0.5}n\int_0^1x^{2n}\ \mathrm dx$

$\displaystyle=\int_0^1\sum_{n=0}^\infty (-1)^n\dbinom{-0.5}nx^{2n}\ \mathrm dx$

$\displaystyle=\int_0^1\sum_{n=0}^\infty\dbinom{-0.5}n(-x^2)^n\ \mathrm dx$

$\displaystyle=\int_0^1\frac1{\sqrt{1-x^2}}\ \mathrm dx$

$\displaystyle=\arcsin1-\arcsin0$

$\displaystyle=\frac\pi2$