Convergent or Divergent?

Ans: divergent.
Let f(x)=Σ(-1)^(i+1)C(n,i)x^i / i , i=1,2,3,..., n
(Note: f(0)=0 and the original series is f(1) )
=> f'(x)= Σ C(n,i)(-x)^(i-1), i=1,2, ~ n
- xf'(x)=Σ C(n, i) (-x)^i, i=1,2, ~n
- xf'(x)= (1-x)^n - 1 => f'(x)= [1- (1-x)^n]/ x
=> f(1)-f(0)=∫[0,1] [ 1- (1-x)^n ]/ x dx
=> f(1) = ∫[0,1] (1- u^n)/(1-u) du (u= 1-x)
= ∫[0,1] [1+u+u^2+...+u^(n-1)] du
= 1+ 1/2+ 1/3+ 1/4 + ... + 1/n
so, Σ(-1)^(i+1) C(n,i)/i = 1+ 1/2+ 1/3+ ... + 1/n -->∞

Note: (1) 1+ 1/2 + 1/3 + ... + 1/n + ... is p-series p=1 => divergent
or (2) 1+ 1/2+ 1/3+....+ 1/n > ∫[2, n+1] 1/x dx = ln(n+1) - ln2 -->∞

good job