# 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.

Hint: Evaluate the series up to a finite $$N$$ first, then take the limit as $$N\to \infty$$. The series up to a finite $$N$$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.
• I see it, I forgot about that series! I will post the solution in a bit but I am seeing that $c=1$?
• Since $\sum\limits_{k=1}^{N}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{N+1}\to 1$, you are right. :) Mar 4, 2019 at 6:32
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.