<Dd> c = 1 + 2 + 3 + 4 + 5 + 6 + ⋯ 4 c = 4 + 8 + 12 + ⋯ c − 4 c = 1 − 2 + 3 − 4 + 5 − 6 + ⋯ (\ displaystyle (\ begin (alignedat) (7) c& () = () &1 + 2&& () + 3 + 4&& () + 5 + 6 + \ cdots \ \ 4c& () = () &4&& () + 8&& () + 12 + \ cdots \ \ c - 4c& () = () &1 - 2&& () + 3 - 4&& () + 5 - 6 + \ cdots \ \ \ end (alignedat))) </Dd> <P> The second key insight is that the alternating series 1 − 2 + 3 − 4 + ⋯ is the formal power series expansion of the function 1 / (1 + x) but with x defined as 1 . Accordingly, Ramanujan writes: </P> <Dl> <Dd> − 3 c = 1 − 2 + 3 − 4 + ⋯ = 1 (1 + 1) 2 = 1 4 (\ displaystyle - 3c = 1 - 2 + 3 - 4 + \ cdots = (\ frac (1) ((1 + 1) ^ (2))) = (\ frac (1) (4))) </Dd> </Dl> <Dd> − 3 c = 1 − 2 + 3 − 4 + ⋯ = 1 (1 + 1) 2 = 1 4 (\ displaystyle - 3c = 1 - 2 + 3 - 4 + \ cdots = (\ frac (1) ((1 + 1) ^ (2))) = (\ frac (1) (4))) </Dd>

If you add every number from 1 to infinity