Moment generating function from probability mass function

Question

We are given the pmf: $$f_X(k) = \frac{1}{k(k+1)}, k \geq 1 $$ and we have to compute the moment generating function. So far I've got: $$M_X(t) = E[e^{tx}] = \sum_{k=1}^{\infty} e^{tk} \frac{1}{k(k+1)}$$ $\sum_{k=1}^{\infty} \frac{1}{k(k+1)}$ is a telescoping sequence right that equals to 1? So how do I move forward from here?

Answer 1

First, assume that $t<0$ since otherwise the series clearly diverges. Then, note that $\sum\limits_{k=0}^\infty e^{tk}=\dfrac{1}{1-e^t}$, by geometric series. Now, consider the integral of both sides over $t$ $$ \begin{align*} \int \sum\limits_{k=0}^\infty e^{tk}\, dt&= \int \dfrac{dt}{1-e^t}\\ t + \sum\limits_{k=1}^\infty \dfrac{e^{tk}}{k}&=-\ln (1-e^t)\\ \sum\limits_{k=1}^\infty \dfrac{e^{tk}}{k}&=-\ln (1-e^t) - t. \end{align*} $$

Now, we unbox your MGF. $$ \begin{align*} M_X(t)= \sum\limits_{k=1}^\infty e^{tk}\dfrac{1}{k(k+1)}&= \sum\limits_{k=1}^\infty e^{tk}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\\ &= \sum\limits_{k=1}^\infty \left(\dfrac{e^{tk}}{k}-\dfrac{e^{tk}}{k+1}\right)\\ &= \sum\limits_{k=1}^\infty \dfrac{e^{tk}}{k}-\frac{1}{e^t}\sum\limits_{k=1}^\infty\dfrac{e^{t(k+1)}}{k+1}\\ &= \sum\limits_{k=1}^\infty \dfrac{e^{tk}}{k}-\frac{1}{e^t}\sum\limits_{k=2}^\infty\dfrac{e^{tk}}{k}\\ &= \left(-\ln(1-e^t)-t \right)-\frac{1}{e^t}\left(-\ln(1-e^t)-t -e^t\right)\\ & = 1+(e^{-t}-1)(\ln(1-e^t)+t). \end{align*} $$