∫n→∞, 1/n (1/3) → 0, ∴ ln [1+1/n (1/3)] ~ 65438+\

But ∑ [(- 1) n]/n (1/3) is a staggered sequence, which satisfies 1/n (1/3) → 0 and 1/n (65438+). 1/(n+1) (1/3), that is, the condition of Leibniz discriminant method, ∴ series convergence. But ∑ [(-1) n]/n (1/3) = ∑1/n (1/3), that is, p = 1/3. P series of 1, divergent. ∴∑ [(- 1) n]/n (1/3) conditional convergence.

∴ convergence of series ∑ [(-1) n] ln [1+1/n (1/3)] is conditional convergence.

For reference.