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Goog Morning Everyone,

I'd like to ask the following exercise that my professor gave,

that i think it has more theory behind it than expected it.

The exercise ask to prove what the following series is equal to :

$$\sum_{k \in \mathbb{N^{+}}} \frac{1}{k(k+1)(k+2) \cdots(k+r)} =?$$

I think the series is very similar to the Mengoli Series, or at least the method used to solve it (I am reffering to the partial fraction decomposition).

I tried with last one but i was unable to correctly solve for (r+1) terms.

I know that Eulero-Beta function can be used too, But i'm not super familiar with it so i'd like to use the partial fraction decomposition.

Anyway, any tips or help would be amazing,

Thank you anyway

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Hint $$\frac{r}{k(k+1)(k+2) \cdots(k+r)}=\frac{1}{k(k+1)(k+2) \cdots(k+r-1)}-\frac{1}{(k+1)(k+2) \cdots(k+r)}$$

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You can use the following decomposition: $$\frac{r}{\prod_{i=0}^r(x+i)}=\frac{1}{\prod_{i=0}^{r-1}(x+i)}-\frac{1}{\prod_{i=1}^r(x+i)} .$$

So, $$\sum_{k>=1}\frac{1}{\prod_{i=0}^{r}(k+i)}=\frac{1}{r}(\sum_{k>=1}\frac{1}{\prod_{i=0}^{r-1}(k+i)}-\sum_{k>=1}\frac{1}{\prod_{i=1}^{r}(k+i)})=\frac{1}{r}(\sum_{k>=1}\frac{1}{\prod_{i=0}^{r-1}(k+i)}-\sum_{k>=2}\frac{1}{\prod_{i=0}^{r-1}(k+i)})=\frac{1}{r}(\frac{1}{\prod_{i=0}^{r-1}(1+i)}+\sum_{k>=2}\frac{1}{\prod_{i=0}^{r-1}(k+i)}-\sum_{k>=2}\frac{1}{\prod_{i=0}^{r-1}(k+i)})=\frac{1}{r*r!}$$

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