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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.

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    $\begingroup$ Hint: $\frac{1}{k+1} = \int_0^1 t^k dt$ $\endgroup$ – achille hui Apr 18 '15 at 17:09
  • $\begingroup$ I am supposed to work this out without calculus, I sadly don't know what that would even mean. $\endgroup$ – John Doe Apr 18 '15 at 17:10
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    $\begingroup$ 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. $\endgroup$ – achille hui Apr 18 '15 at 17:14
  • $\begingroup$ Will pay attention to that next time. Appreciated anyhow :) $\endgroup$ – John Doe Apr 18 '15 at 17:16
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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 $$\frac{1}{20}\sum_{k=1}^{20}(1)^{20-k}\left(-1\right)^{k+1}{20\choose k}=\frac{1}{20}\left[1-\sum_{k=0}^{20}(1)^{20-k}\left(-1\right)^k{20\choose k}\right]=\frac{1}{20}\left[1-(1-1)^{20}\right]=\frac{1}{20}$$

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    $\begingroup$ Equal to? Not quite! $\endgroup$ – Simon S Apr 18 '15 at 17:19
  • $\begingroup$ @SimonS: Thanks, I had a fault. $\endgroup$ – Ángel Mario Gallegos Apr 18 '15 at 17:38
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$$\frac{1}{k+1}\binom{n}{k}=\frac{n!}{(n-k)!(k+1)!}=\frac1{n+1}{n+1\choose k+1}$$

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