Harmonic series in probability mass function problem Suppose $X$ is a discrete random variable with possible values $\{1, 2, 3,\dots\}.$ Further, suppose the p.m.f is $$c\left(\frac{1}{x}-\frac{1}{x+1}\right)\enspace\text{s.t. $c > 0$}$$
Find c and $E[X].$
Idea:
We have $$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\left(\sum_{x=1}^{\infty}\frac{1}{x}-\sum_{x=1}^{\infty}\frac{1}{x+1}\right)$$
But since $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series, diverges. Thus, there is no value for $c$.
Since it diverges, $E[X]$ does not exist.
Questions:
Is it possible for c not to exist? Did I do a mistake?
Update:
$$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\sum_{x=1}^{\infty}\left(\frac{1}{x}-\frac{1}{x+1}\right)$$
by telescoping series we have 
$$1=c\cdot 1$$
So, our p.m.f is $$\left(\frac{1}{x}-\frac{1}{x+1}\right)$$
But, $$E[X]=1\cdot\frac{1}{2}+2\cdot\frac{1}{6}+3\cdot\frac{1}{12}+4\cdot \frac{1}{20}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots$$
But that diverges. So, $E[X]$ doesn't exist.

You can write $$\sum a_n-b_n=\sum a_n-\sum b_n$$only if at least one of $\sum a_n$ or $\sum b_n$ is bounded. In this case$$\sum_{x=1}^{\infty}{1\over x}-{1\over x+1}{=\left(1-{1\over 2}\right)\\+\left({1\over 2}-{1\over 3}\right)\\+\left({1\over 3}-{1\over 4}\right)\\+\left({1\over 4}-{1\over 5}\right)\\+\cdots\\=1}$$therefore $$c=1$$ and the rest is easy.
