# Value of the sum: $\binom{19}{0} - 1/2\binom{19}{1} +1/3\binom{19}{2} - 1/4\binom{19}{3} … -1/20\binom{19}{19}$?

How do I find the value of this sum? I tried taking out the equal binomial coefficients as factors but this didn't really simplify anything. I am stumped.

• Hint: $\frac{1}{k+1} = \int_0^1 t^k dt$ – achille hui Apr 18 '15 at 17:09
• I am supposed to work this out without calculus, I sadly don't know what that would even mean. – John Doe Apr 18 '15 at 17:10
• Calculus way: $$\sum_{k=0}^{19} \frac{(-1)^k}{k+1}\binom{19}{k} = \int_0^1 \sum_{k=0}^{19} \binom{19}{k} (-t)^k dt = \int_0^1 (1-t)^{19} dt = \frac{1}{20}$$ If you are not allowed to use calculus, you should state that in your question. – achille hui Apr 18 '15 at 17:14
• Will pay attention to that next time. Appreciated anyhow :) – John Doe Apr 18 '15 at 17:16

Observe $$\frac{1}{k} {{n-1}\choose k-1}=\frac{1}{k}\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{1}{n}\frac{n(n-1)!}{k!(n-k)!}=\frac{1}{n}{n\choose k}$$ The sum you are looking for is equal, according to binomial theorem, to $$\frac{1}{20}\sum_{k=1}^{20}(1)^{20-k}\left(-1\right)^{k+1}{20\choose k}=\frac{1}{20}\left[1-\sum_{k=0}^{20}(1)^{20-k}\left(-1\right)^k{20\choose k}\right]=\frac{1}{20}\left[1-(1-1)^{20}\right]=\frac{1}{20}$$
$$\frac{1}{k+1}\binom{n}{k}=\frac{n!}{(n-k)!(k+1)!}=\frac1{n+1}{n+1\choose k+1}$$