Can anyone help me ?

wassimamri said:
the question is to find the value of this

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Assume p=3 and write out the expanded summation and multiplication. Work with inner multiplication first leaving 'k' as 'k'. Do you see anything special developing?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
i tried to decompose into partial fractions
 

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wassimamri said:
nothing
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Since you know nothing about Gamma, and Beta I'll share my solution using them knowing you won't understand. Partial fractions might work but seems tedious. Perhaps somebody can make use of this and guide you with the partial fractions.
∑k=1∞1∏j=0p(k+j)=∑k=1∞Γ(k)Γ(p+k+1) (Rewrite in terms of Gamma function)\displaystyle \sum_{k=1}^{\infty}\frac{1}{\prod_{j = 0}^p (k + j)}= \sum_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(p+k+1)}\text{ (Rewrite in terms of Gamma function)}

=1Γ(p+1)∑k=1∞Γ(k)Γ(p+1)Γ(p+k+1) (Complete the Beta function)\displaystyle =\frac{1}{\Gamma(p+1)}\sum_{k=1}^{\infty}\frac{\Gamma(k)\Gamma(p+1)}{\Gamma(p+k+1)} \text{ (Complete the Beta function)}

=1Γ(p+1)∑k=1∞B(p+1,k)\displaystyle =\frac{1}{\Gamma(p+1)} \sum_{k=1}^{\infty}\Beta(p+1,k)

=1Γ(p+1)∑k=1∞∫01αk−1(1−α)p dα\displaystyle =\frac{1}{\Gamma(p+1)} \sum_{k=1}^{\infty} \int_{0}^{1} \alpha^{k-1}(1-\alpha)^{p}\, d\alpha

=1Γ(p+1)∫01∑k=1∞αk−1(1−α)p dα\displaystyle =\frac{1}{\Gamma(p+1)}\int_{0}^{1}\sum_{k=1}^{\infty} \alpha^{k-1}(1-\alpha)^{p}\, d\alpha

=1Γ(p+1)∫01(1−α)p−1 dα\displaystyle =\frac{1}{\Gamma(p+1)}\int_{0}^{1}(1-\alpha)^{p-1}\, d\alpha

=1Γ(p+1)⋅1p=1p!×p\displaystyle =\frac{1}{\Gamma(p+1)}\cdot \frac{1}{p}= \frac{1}{p!\times p}

where Γ()\Gamma() is the Gamma function and B()\Beta() is the Beta function.
 
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