How to find a moment generating function given $p(x)=\frac1{x(x+1)}$ for $x\geqslant1$?

Question

It was given to find the moment generating function for $p(x)=\frac1{x(x+1)}$ for $x\geqslant1$. So i tried doing E(e^(t*x)) and ended up in summation of 1+(e^t/2)+((e^(2*t)/6)+((e^(3*t)/12)+.....Now i don't know what to do.I can't sum up this sequence.

Answer 1

First note that $p_n=\frac1n - \frac1{n+1}$ hence $p_n = \frac1{n(n+1)}$ is the probability mass function of a random variable $X$. Now if $|s|<1$, we have $$\mathbb E\left[|s^X|\right] = \sum_{n=1}^\infty \frac{|s|^n}{n(n+1)}\leqslant\sum_{n=1}^\infty \frac1{n(n+1)}=1, $$ and so by absolute convergence of the above series, the generating function of $X$ is \begin{align} \mathbb E\left[s^X\right] &= \sum_{n=1}^\infty s^n\left(\frac1n-\frac1{n+1}\right)\\ &= \sum_{n=1}^\infty \frac{s^n}n - \sum_{n=1}^\infty \frac{s^n}{n+1}\\ &= -\log(1-s) - \frac1s\left(\sum_{n=2}^\infty \frac{s^n}n\right)\\ &= -\log(1-s) -\frac1s\left(-\log(1-s)-s\right)\\ &= 1+\frac1s(1-s)\log(1-s). \end{align} The moment generating function would likewise be given by $$\mathbb E\left[e^{tX}\right] = 1+\frac1{e^t}(1-e^t)\log(1-e^t)\qquad (t<0). $$