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$$\frac{\binom{b+i}{1}}{\binom{i}{1}} + \frac{\binom{b+i}{2}}{\binom{i}{2}} + \frac{\binom{b+i}{3}}{\binom{i}{3}} +... + \frac{\binom{b+i}{i}}{\binom{i}{i}} $$

where $b \in \mathbb{R}$ and $b\in [-1, 1]$ .

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    @ThomasAndrews I thought that the series had combinatorials similar to that of the power series. I therefore used the title. Changed it. – 2017-02-02

2 Answers 2

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Maple says it is $$ \frac{b+i}{1-b} - \frac{\Gamma(b+i+1)}{i!(1-b) \Gamma(b)} $$ (assuming of course $b \ne 1$).

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    I tried substituting values of $b$ and $i$, it works for the initial substitution, but fails when I substitute $i=2$. – 2017-02-02
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    @learner: You made somewhere a mistake. – 2017-02-02
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    @user90369 Sorry for the mistake, I got it. Thanks so much for the vivid explanation. – 2017-02-02
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    @learner: You are welcome! – 2017-02-02
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Proof of $ \displaystyle\enspace \sum\limits_{j=1}^i \binom{b+i}{j}\binom{i}{j}^{-1} = \frac{b+i}{1-b}-\frac{b}{1-b} \binom{b+i}{i} \enspace$ by induction for $b\ne 1$.

$i:=1$ : $\displaystyle \enspace \sum\limits_{j=1}^1 \binom{b+1}{j}\binom{1}{j}^{-1}=\frac{b+1}{1-b}-\frac{b}{1-b} \binom{b+1}{1} \enspace $ which is true because of $b+1=b+1$

$i\to i+1$ :

We have $\displaystyle \enspace \sum\limits_{j=1}^{i+1} \binom{b+i+1}{j}\binom{i+1}{j}^{-1} =\frac{b+i+1}{i+1}(1+\sum\limits_{j=1}^i \binom{b+i}{j}\binom{i}{j}^{-1}) \enspace $ and therefore have to proof $\displaystyle \enspace \frac{b+i+1}{1-b}-\frac{b}{1-b} \binom{b+i+1}{i+1} =\frac{b+i+1}{i+1}(1+\frac{b+i}{1-b}-\frac{b}{1-b} \binom{b+i}{i}) $ .

But elementary transformations show immediately, that this is true.

A note for $\enspace b=1$ :

$ \displaystyle\enspace \sum\limits_{j=1}^i \binom{1+i}{j}\binom{i}{j}^{-1}=(i+1)H_i \enspace $ where $H_i$ is the $i^{th}$ partial sum of the harmonic series.