A coin is tossed until the first time a head turns up. If this occurs on the n th toss and n is odd you win 2n/n , but if n is even then you lose 2n/n . Then if your expected winnings exist they are given by the convergent series.

Answer:This is a series of variable harmonics that converges to log (2).Step-by-step explanation:Let's ignore for a moment that n is even or odd, and look at the expected value for any n. Let X be profit, which can be negative if we lose. The wait is given using (and assuming the coin is valid) E[X]=∑n=1∞((−1)n+1×2∧n/n)⋅1/2∧n=∑n=1∞(−1)n+1×1/n.This is a series of variable harmonics that converges to log (2). However, expectation exists only if it absolutely converges! Looking at ∑n=1∞∣(−1)n+1×2∧n/n∣×1/2n=∑n=1∞1/nwe notice that a number of harmonics diverge, therefore, in fact, there is no expectation of X.