Odd number of heads from increasingly biased coins

You have $n$ coins. Coin $k$ (for $k=1,\dots,n$) lands heads with probability $\tfrac{1}{2k+1}$. All $n$ coins are tossed once. What is the probability that the total number of heads is odd?

$\tfrac{n}{2n+1}$. For independent coins, $P(\text{even})-P(\text{odd})=\prod_k(1-2p_k)=\prod_{k=1}^n\tfrac{2k-1}{2k+1}=\tfrac{1}{2n+1}$ by telescoping. So $P(\text{odd})=\tfrac12\left(1-\tfrac{1}{2n+1}\right)=\tfrac{n}{2n+1}$.