IMC 2010 · Day 1 · P2

Let $$F(x) = \sum_{k=0}^{\infty} \frac{x^{4k+4}}{(4k+1)(4k+2)(4k+3)(4k+4)}.$$ This power series converges for $|x| \le 1$ and our goal is to compute $F(1)$.

Differentiating 4 times, we get $$F^{(IV)}(x) = \sum_{k=0}^{\infty} x^{4k} = \frac{1}{1 - x^4}.$$ Since $F(0) = F'(0) = F''(0) = F'''(0) = 0$ and $F$ is continuous at $1 - 0$ by Abel's continuity theorem, integrating 4 times we get $$F'''(y) = F'''(0) + \int_0^y F^{(IV)}(x)\,dx = \int_0^y \frac{dx}{1 - x^4} = \frac{1}{2} \arctan y + \frac{1}{4} \log(1+y) - \frac{1}{4} \log(1-y),$$ $$\begin{align*} F''(z) &= F''(0) + \int_0^z F'''(y)\,dx = \int_0^z \left( \frac{1}{2} \arctan y + \frac{1}{4} \log(1+y) - \frac{1}{4} \log(1-y) \right) dy = \\ &= \frac{1}{2} \left( z \arctan z - \int_0^z \frac{y}{1+y^2}\,dy \right) + \frac{1}{4} \left( (1+z)\log(1+z) - \int_0^z dy \right) + \frac{1}{4} \left( (1-z)\log(1-z) + \int_0^z dy \right) = \\ &= \frac{1}{2} z \arctan z - \frac{1}{4} \log(1+z^2) + \frac{1}{4} (1+z)\log(1+z) + \frac{1}{4} (1-z)\log(1-z), \end{align*}$$ $$\begin{align*} F'(t) &= \int_0^t \left( \frac{1}{2} z \arctan z - \frac{1}{4} \log(1+z^2) + \frac{1}{4} (1+z)\log(1+z) + \frac{1}{4} (1-z)\log(1-z) \right) dt = \\ &= \frac{1}{4} \left( (1+t^2)\arctan t - t \right) - \frac{1}{4} \left( t \log(1+t^2) - 2t + 2\arctan t \right) + \\ &\quad + \frac{1}{8} \left( (1+t)^2 \log(1+t) - t - \frac{1}{2} t^2 \right) - \frac{1}{8} \left( -(1-t)^2 \log(1-t) + t - \frac{1}{2} t^2 \right) = \\ &= \frac{1}{4} (-1 + t^2) \arctan t - \frac{1}{4} t \log(1+t^2) + \frac{1}{8} (1+t)^2 \log(1+t) - \frac{1}{8} (1-t)^2 \log(1-t), \end{align*}$$ $$\begin{align*} F(1) &= \int_0^1 \left( \frac{1}{4} (-1 + t^2) \arctan t - \frac{1}{4} t \log(1+t^2) + \frac{1}{8} (1+t)^2 \log(1+t) - \frac{1}{8} (1-t)^2 \log(1-t) \right) dt = \\ &= \left[ \frac{-3t + t^3}{12} \arctan t + \frac{1 - 3t^2}{24} \log(1+t^2) + \frac{(1+t)^3}{24} \log(1+t) + \frac{(1-t)^3}{24} \log(1-t) \right]_0^1 = \frac{\ln 2}{4} - \frac{\pi}{24}. \end{align*}$$ Remark. The computation can be shorter if we change the order of integrations. $$\begin{align*} F(1) &= \int_{t=0}^{1} \int_{z=0}^{t} \int_{y=0}^{z} \int_{x=0}^{y} \frac{1}{1-x^4}\,dx\,dy\,dz\,dt = \int_{x=0}^{1} \frac{1}{1-x^4} \left( \int_{y=x}^{1} \int_{z=y}^{1} \int_{t=z}^{1} dt\,dz\,dy \right) dx = \\ &= \int_{x=0}^{1} \frac{1}{1-x^4} \cdot \frac{(1-x)^3}{6}\,dx = \left[ -\frac{1}{6} \arctan x - \frac{1}{12} \log(1+x^2) + \frac{1}{3} \log(1+x) \right]_0^1 = \frac{\ln 2}{4} - \frac{\pi}{24}. \end{align*}$$

Let $$A_m = \sum_{k=0}^{m} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \sum_{k=0}^{m} \left( \frac{1}{6} \cdot \frac{1}{4k+1} - \frac{1}{2} \cdot \frac{1}{4k+2} + \frac{1}{2} \cdot \frac{1}{4k+3} - \frac{1}{6} \cdot \frac{1}{4k+4} \right),$$ $$B_m = \sum_{k=0}^{m} \left( \frac{1}{4k+1} - \frac{1}{4k+3} \right),$$ $$C_m = \sum_{k=0}^{m} \left( \frac{1}{4k+1} - \frac{1}{4k+2} + \frac{1}{4k+3} - \frac{1}{4k+4} \right) \quad \text{and}$$ $$D_m = \sum_{k=0}^{m} \left( \frac{1}{4k+2} - \frac{1}{4k+4} \right).$$ It is easy check that $$A_m = \frac{1}{3} C_m - \frac{1}{6} B_m - \frac{1}{6} D_m.$$ Therefore, $$\lim A_m = \lim \frac{2 C_m - B_m - D_m}{6} = \frac{2 \ln 2 - \frac{\pi}{4} - \frac{1}{2} \ln 2}{6} = \frac{\ln 2}{4} - \frac{\pi}{24}.$$