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In order to continue solving a probability related exercise, I have to extract the value of the constant $c$ from the following: (while $x\in \mathbb{N}$) $$\sum_{x=1}^{\infty}\frac{c}{x(x+1)(x+3)}=1$$

My first step was to decompose the above to the partial fractions: $c \cdot \sum_{x=1}^{\infty}{((\frac{1}{3x})-(\frac{1}{2(x+1)})+(\frac{1}{6(x+3)}))} = 1$

Then, tried to combine elements of the series to see if I can arrange them in a telescopic way but that failed as well.

Could you please provide a hint or a way to approach this sum calculation?

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    prove that your sum is equal to $$\frac{1}{4}c$$ – 2017-01-07
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    @Dr.SonnhardGraubner I fixed a typing mistake. made the denominator x(x+1)(x+3). What should be the sum, please? – 2017-01-07
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    @KenMizrahi The new sum is $\frac{7c}{36}$. – 2017-01-07
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    yes this is true – 2017-01-07

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Hint. Observe that $$ \begin{align} \frac{3}{x(x+1)(x+3)}&=\left(\frac{1}{x}-\frac{1}{x+1}\right)-\left(\frac{1}{2(x+1)}-\frac{1}{2(x+2)}\right)-\left(\frac{1}{2(x+2)}-\frac{1}{2(x+3)}\right) \end{align} $$ then terms telescope nicely.

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    I fixed a typing mistake in the question. Is there a similar equation for the case with (x+3) instead of (x+2)? – 2017-01-07
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    thank you! If I got you right, I think there should be a plus sign for the last element. Right? – 2017-01-07