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?

  • $\begingroup$ $\sum_{i=0}^j (-1)^i\frac{C_i}{2i}$ 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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.