∑i=1∞(1n)i={1n−1(n≥2)∞(n=1)−∞(n=−1)1n+1(n≤−2)\sum^{\infty }_{i=1}{(\frac{1}{n})^i}=\left\{ \begin{matrix} \frac{1}{n-1} (n\ge 2 ) \\ \infty (n=1) \\ -\infty (n=-1) \\ \frac{1}{n+1} (n\le-2) \end{matrix}\right.∑i=1∞(n1)i=⎩⎨⎧n−11(n≥2)∞(n=1)−∞(n=−1)n+11(n≤−2)
