Show $P(X\ge k)=c/k$
I tried getting the integral for $X\ge k$. Didn't work out.
$$\int_k^{\infty} c/k(k+1)dk=-c*ln(|(1/\infty)+1|)+c*ln(|(1/k)+1|)$$
Which equals $c*ln(|(1/k)+1|)$, which I don't think can equal c/k
Find c
$$\int_1^\infty c/k(k+1)dk=1$$
So
$$-c*ln(|(1/\infty)+1|)+c*ln(|(1+1|)=1$$
Since $ln(|(1/+-\infty)+1|)=0$ I'm guessing c=1/ln(2)?
Find the probability that X is odd
For this I know that since this is a RV than spans from 1 to $\infty$ then probably half of the values will be odd. I'm guessing take the integral from 0 to infinity for c/(2k+1)(2k+2)?
HINT
$X$ is an integer-valued random variable! And $P(X=x)$ is an actual probability, not a probability density (pdf). You cannot integrate $P()$ (you can only integrate pdf's). Instead you need something easier - you can add them.
$$P(X \ge k) = P(X=k) + P(X=k+1) + P(X=k+2) + \dots = \sum_{j=k}^\infty P(X=j)$$
So you need to prove the above infinite summation equals $c/k$.
Finding $c$ is really easy: $P(X \ge 1) = 1$ by definition of $X$, and you already know what $P(X \ge k)$ is...
$P(X \,\,\text{odd})$ is yet another exercise in infinite summation.