Show that $ \mathbb{E} < \infty \Longleftrightarrow \sum_{n=1}^\infty \mathbb{P} < \infty $ for random variable $X$

Let $X$ be a discrete random variable taking values in $\mathbb{N}$.

Show that
$
\mathbb{E}[X] < \infty \Longleftrightarrow \sum_{n=1}^\infty \mathbb{P}[X>n] < \infty
$.

I am not that good at math, but could you please explain to me why this does not work the way I think it does .. I am going to speak loud such that you can follow my thoughts.

This is a sum over all probabilities for $ X > n $:

$ \sum_{n=1}^\infty \mathbb{P}[X>n] $

This is the general definition of the expected value:

$ \mathbb{E}[X] = \sum_{i=1}^\infty x_ip_i $

a weighted sum over all probabilites of a RV $X$.

Therefore:

$\mathbb{E}[x] = n_1 \mathbb{P}[X > n_1] + n_2 \mathbb{P}[X > n_2] + … + n_\infty \mathbb{P}[X > n_\infty] $

And this is what I want to know if this would make sense:

Because $ n_1 \mathbb{P}[X > n_1] = n_1 \sum_{j=1}^\infty \mathbb{P}[([X= j + n_1] $
we can write:

\begin{align}
\mathbb{E}[x] &= n_1 \sum_{j=1}^\infty \mathbb{P}[([X= j + n_1] + n_2 \sum_{j=1}^\infty \mathbb{P}[([X= j + n_2] + .. + n_\infty \sum_{j=1}^\infty \mathbb{P}[([X= j + n_\infty] \\
&= \sum_{n=1}^\infty \sum_{j=1}^\infty \mathbb{P}[X = j + n] \\
&= \sum_{n=1}^\infty \sum_{j=n+1}^\infty \mathbb{P}[X = j] \\
&= \sum_{j=2}^\infty \sum_{n=1}^j \mathbb{P}[X = j] \\
&= \sum_{j=2}^\infty j * \mathbb{P}[X = j] = \mathbb{E}[X]
\end{align}

And this makes sense in my eyes since it is given that $ X > n$ and therefore $\mathbb{E}[X]$ must start with $j=2$ here. Unfortunatelly I was told that I can not write this since I do not follow the definition of the expected value whereas I think that I do.

Can anyone explain to me where I am making the mistake?

Solutions Collecting From Web of "Show that $ \mathbb{E} < \infty \Longleftrightarrow \sum_{n=1}^\infty \mathbb{P} < \infty $ for random variable $X$"

Since $X$ takes values in $\mathbb{N}$,
$$
\begin{align}
\mathrm{E}[X]
&=\sum_{k=0}^\infty(k+1)\overbrace{(\mathrm{P}[X\gt k]-\mathrm{P}[X\gt k+1])}^{\mathrm{P}[X=k+1]}\\
&=\lim_{N\to\infty}\left(\sum_{k=0}^N(k+1)\mathrm{P}[X\gt k]-\sum_{k=0}^{N+1} k\mathrm{P}[X\gt k]\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^N\mathrm{P}[X\gt k]-\color{#C00000}{\lim_{N\to\infty}(N+1)\mathrm{P}[X\gt N+1]}\\
&=\sum_{k=0}^\infty\mathrm{P}[X\gt k]
\end{align}
$$
If $\mathrm{E}[X]\lt\infty$, then the sum for $\mathrm{E}[X]$ converges, and therefore, its tail must vanish
$$
\begin{align}
0
&=\lim_{N\to\infty}\sum_{k=N}^\infty(k+1)(\mathrm{P}[X\gt k]-\mathrm{P}[X\gt k+1])\\
&\ge\lim_{N\to\infty}(N+1)\sum_{k=N}^\infty(\mathrm{P}[X\gt k]-\mathrm{P}[X\gt k+1])\\[6pt]
&=\lim_{N\to\infty}(N+1)\mathrm{P}[X\gt N]\\[10pt]
&\ge\color{#C00000}{\lim_{N\to\infty}(N+1)\mathrm{P}[X\gt N+1]}
\end{align}
$$

On the wikipedia article on expectation you can find a proof of $E(X) = \sum_{n=1}^{\infty} P(X>n)$. From here it follows immediately that one of these expressions is finite if and only if the other one is.