Bernoulli Number Sum: At B4 i get 1/180?

Question

I know i did something wrong because i don't get -1/30. I was able to get zero for B3 and 1/6 for B2, but then

$$B_4=1-{4\choose 0}\frac{B_0}{4-0+1}-{4\choose 1}\frac{B_1}{4-1+1}-{4\choose 2}\frac{B_2}{4-2+1}-{4\choose 3}\frac{B_3}{4-3+1}=$$

$$B_4=1-\frac{4}{(4-0)!0!}\frac{B_0=1}{4-0+1}-\frac{4}{(4-1)!1!}\frac{B_1=-1/2}{4-1+1}-\frac{4}{(4-2)!2!}\frac{B_2=1/6}{4-2+1}-\frac{4}{(4-3)!3!}\frac{B_3=0}{4-3+1}=$$

$$B_4=1-\frac{4}{24}\frac{1}{5}-\frac{4}{6}\frac{-1/2}{4}-\frac{4}{4}\frac{1/6}{3}-\frac{4}{6}\frac{0}{2}= $$

$$B_4=1-\frac{4}{120}-\frac{-2}{24}-\frac{1}{18}-\frac{0}{12}= $$

$$B_4=1-\frac{1}{30}-\frac{-1}{12}-\frac{1}{18}-0= $$

$$B_4=1-\frac{6}{180}-\frac{-15}{180}-\frac{10}{180}-0 \not= \frac{-1}{30} $$

i think im failing to be recursive somewhere

Answer 1

You wrote $4$ where you should have $4!$ from the binomial coefficients. That explains the factor $6$. Also, you're using the recursion for the $+$ convention but substituting the values for the $-$ convention. You should either use $B_2=+\frac12$ or drop the initial $1$ in the recursion.

Answer 2

You made a mistake with the binomial coefficient and used the wrong recursion formula (see https://en.wikipedia.org/wiki/Bernoulli_number#Recursive_definition for $B_m^{-}$). For the variant with $B_1=-1/2$ you get

$$B_4= -\sum_{k=0}^3 {4 \choose k} \frac{B_k}{5-k}$$ $$B_4= -\left(1\frac{1}{5} + 4\frac{-1}{2}\frac{1}{4} + 6\frac{1}{6}\frac{1}{3}\right) = -\frac{1}{30} $$