∀n∈ℕ:∑(-1)^(k+1)(1/k) [from k=1 to 2n] = ∑1/(n+k) [from k=1 to n]
Prove by mathematical induction that $\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $