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So there was this binomial series problem I ran in to while solving some probability problems, and I ended up getting the expression of the sum:

$$\frac{C_0}{j}-\frac{C_1}{j+1}+\frac{C_2}{j+2}-....+(-1)^j\frac{C_j}{2j}=\sum_{i=0}^j (-1)^i\frac{C_i}{2i} $$

My solution to this series: Consider the following integral: $$ \int_0^1 x^{j-1}(1-x)^j\, dx=\int_0^1 (C_0x^{j-1}-C_1x^j+...+(-1)^jC_jx^{2j-1})dx$$ $$=\frac{C_0}{j}-\frac{C_1}{j+1}+\frac{C_2}{j+2}-....+(-1)^j\frac{C_j}{2j}$$

Next to solve this integral, consider another integral:

$$\int_0^1 [\lambda x +(1-x)]^{2j-1}\, dx=\int_0^1 [\lambda x +(1-x)]^{2j-1}\, dx=$$$$\int_0^1 [1-(1-x)\lambda )]^{2j-1}\, dx=\frac{1}{2j}\frac{\lambda ^{2j}-1}{\lambda -1}=\frac{1}{2j} (\lambda ^{2j-1}+\lambda ^{2j-2}+...+\lambda +1)$$ $$\int_0^1 [\lambda x +(1-x)]^{2j-1}\, dx =\int_0^1 \Big[ \sum_{k=0}^{2j-1} \binom{2j-1}{k}(\lambda x)^{2j-1-k}(1-x)^k \Big]\, dx$$ $$= \sum_{k=0}^{2j-1} \Big[ \binom{2j-1}{k}\lambda ^{2j-1-k}\int_0^1 x^{2j-1-k}(1-x)^k \, dx\Big]$$

This is all I have so far, and stuck here to be honest, but is this correct? Because I just assumed that the integral can go inside the summation without even testing for uniform convergence of the series. Can anyone tell me if I am going in the right direction?

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  • $\begingroup$ $\sum_{i=0}^j (-1)^i\frac{C_i}{2i}$ can't be because the term for $i=0$ diverges. $\endgroup$
    – Andreas
    Nov 26, 2017 at 16:36
  • $\begingroup$ Ok I am quite confused, because it diverges this summation doesnt exist to begin with? $\endgroup$ Nov 26, 2017 at 17:26
  • $\begingroup$ The denominator is i+j, not 2i. $\endgroup$ Nov 26, 2017 at 18:24

1 Answer 1

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$$\int_0^1 x^{i-1} (1-x)^i dx =B (i,i+1)=\frac{\Gamma (i)\Gamma (i+1)}{\Gamma (2i+1) }=\frac{(i-1)! i!}{(2i)!}=\frac{1}{ i{2i\choose i }}$$

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