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.)

\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}
