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