可以用D'Alembert比值判别法.a[n] = 1/n²,a[n+1] = 1/(n+1)²,因此a[n+1]/a[n] → 1.对z ≠ 0,a[n+1]·z^(n+1)/(a[n]·z^n) → z.故级数∑{1 ≤ n} z^n/n² = ∑{1 ≤ n} a[n]·z^n在|z| < 1时收敛,|z|...