Proving that $E[X]= \sum_{k=0}^{\infty} P(X>k)$ by proving $(n+1)P(X>n) \xrightarrow[]{n \to \infty} 0$

Question

Let $X$ be a random variable with positive integer range and finite mean. To show that $$E[X]= \sum_{k =0}^{\infty} P(X>k).$$

Proof: I showed using induction that $$\sum_{k=0}^n P(X>k) = \sum_{t=1}^n (t \cdot P(X=t)) + (n+1)P(X>n)$$ If we apply $n \to \infty$, we would get the required result if if we show that as $n \to \infty$, $(n+1)P(X>n)$ tends to $0$. I'm having difficulty proving that.

I know that $n \cdot P(X=n)$ tends to $0$ as $n \to \infty$ because $X$ has a finite mean.

(PS: I do know this question is duplicate but in the previous question I didn't find any proof resembling this. Also to show that $(n+1)P(X>n)$ tends to 0 can be a separate question by itself.)

Answer 1

Note that

$$n \mathbb{P}(X>n) = n \int_{\{X>n\}} \, d\mathbb{P} \leq n \int_{\{X>n\}} \frac{X}{n} \, d\mathbb{P}.$$

Since $X$ is intgrable, we can apply the dominated convergence theorem to conlude

$$\lim_{n \to \infty} (n \mathbb{P}(X>n)) \leq \lim_{n \to \infty} \int_{\{X>n\}} X \, d\mathbb{P}=0.$$

Finally, using that

$$\lim_{n \to \infty} \frac{n+1}{n} = 1$$

we get

$$\lim_{n \to \infty} (n+1) \mathbb{P}(X>n)=0.$$

Remark: The above proof does not only work for integer-valued integrable random variables, but for any non-negative integrable random variable. For the particular case of integer-valued random variables, the proof can be reformulated as follows:

Note that

$$n \mathbb{P}(X>n) = n \sum_{k=n+1}^{\infty} \mathbb{P}(X=k) \leq n \sum_{k=n+1}^{\infty} \frac{k}{n} \mathbb{P}(X=k).$$

Since the expectation $\mathbb{E}(X) = \sum_{k=0}^{\infty} k \mathbb{P}(X=k)$ is finite, we get

$$\lim_{n \to \infty} n \mathbb{P}(X>n) \leq \lim_{n \to \infty} \sum_{k=n+1}^{\infty} k \mathbb{P}(X=k)=0.$$

Answer 2

\begin{align} E[X] &= \sum_{k=1}^\infty k\cdot P(X=k)\\ &= \quad P(X=1)\\ &\qquad + P(X=2) + P(X=2)\\ &\qquad + P(X=3) + P(X=3) + P(X=3)\\ &\qquad + P(X=4) + P(X=4) + P(X=4) + P(X=4)\\ &\qquad\vdots\\ &= P(X > 0) + P(X >1) + P(X>2) + P(X >3) + \cdots & \text{add each column}\\ &= \sum_{k=1}^\infty P(X \ge k). \end{align}