Harmonic series times binomial

Question

I'm solving a question and get the following formula: $\sum_{i=0}^{n}\frac{1}{i+1}\binom{n}{i}\sigma^{i}(1-\sigma)^{n-i}$

I'm curious if there is a simpler form of this formula. I've found some related questions, but still can't solve this one.

Thank you for your response.

Answer 1

Using that $\frac{1}{i+1}=\int\limits_0^1 t^i dt$ and replace the order of the summation and integration we get:

$(1-\sigma)^n\int\limits_0^1\sum\limits_{i=0}^{n}\binom{n}{i}\big(\frac{\sigma t}{1-\sigma}\big)^{i}dt$

With the application of the binomial theorem and perform the integration we have the result:

$(1-\sigma)^n\int\limits_0^1\big(1+\frac{\sigma t}{1-\sigma}\big)^ndt=\int\limits_0^1(1-\sigma+\sigma t)^ndt=\frac{1-(1-\sigma)^{n+1}}{(1+n)\sigma}$

Answer 2

Another approach with calculus is to rewrite $\frac{\sigma^i}{i+1}$ as $\sigma^{-1}\int_0^\sigma x^i dx$ so the sum is $$\sigma^{-1}(1-\sigma)^n\int_0^\sigma \sum_{i=0}^n\binom{n}{i}\left(\frac{x}{1-\sigma}\right)^i dx=\sigma^{-1}\int_0^\sigma\left(1-\sigma+x\right)^ndx=\frac{1-(1-\sigma)^{n+1}}{(n+1)\sigma}.$$

Answer 3

Let us do it a bit differently, use $$\frac{1}{i+1} {n \choose i} =\frac{1}{n+1} { n+1 \choose i+1}.$$ as suggested by abiessu. So the sum $$S=\sum_{i=0}^{n} \frac{1}{i+1} {n \choose i} \sigma^i (1-\sigma)^{n-i} = \frac{1}{n+1} \sum _{i=0}^{n} {n+1 \choose i+1} \sigma^i (1- \sigma)^{n-i}$$ $$\Rightarrow S= \frac{1}{n+1} \sum_{j=1}^{n+1} {n+1 \choose j} ~ \sigma^{j-1} ~ (1-\sigma)^{n+1-j}= \frac{[(\sigma +1 -\sigma)^{n+1}]-(1-\sigma)^{n+1}}{(n+1)\sigma}.$$ Finally $$ S=\frac{1-(1-\sigma)^{n+1}}{(n+1)\sigma}.$$