∀n∈ℕ:∑(-1)^(k+1)(1/k) [from k=1 to 2n] = ∑1/(n+k) [from k=1 to n]

Question

Prove by induction: $$ \sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{n+k} $$

I start with (using $n = t+1$): $$ \sum_{k=1}^{2t+2}(-1)^{k+1}\frac{1}{k}=... $$

but after expanding and attempting to simplify I get stuck at: $$ \sum_{k=1}^{t}\frac{1}{t+k} + (-1)^{2t}\frac{1}{2t+2}-(-1)^{2t}\frac{1}{2t+1} $$

Any suggestions would be very helpful.