Evaluate $\frac{1}{1 \cdot 2 \cdot 3}+\frac{2}{4 \cdot 5 \cdot 6}+\frac{3}{7 \cdot 8 \cdot 9}+ \cdots$

Question

Evaluate $$\frac{1}{1 \cdot 2 \cdot 3}+\frac{2}{4 \cdot 5 \cdot 6}+\frac{3}{7 \cdot 8 \cdot 9}+ \cdots$$ I see this is the same as $$\frac{1}{3} \sum_{i=0}^{\infty} \frac{1}{(3i+1)(3i+2)} $$ $$\frac{1}{3} \sum_{i=0}^{\infty} \frac{1}{(3i+1)} - \frac{1}{(3i+2)} $$ Here I am confused. Any suggestions?

Answer 1

With $\frac1n=\int_0^1x^{n-1}dx$ your sum is$$\frac13\sum_{i\ge0}\int_0^1x^{3i}(1-x)dx=\frac13\int_0^1\frac{1-x}{1-x^3}dx=\frac13\int_0^1\frac{1}{1+x+x^2}dx=\frac{\pi}{9\sqrt{3}}.$$