Let $X$ be a random variable such that $X(\Omega)\subset \mathbb{N}$ in the problem it is said to show that $$ \sum_{k=0}^{n}kP(X=k)=\sum_{k=1}^{n}P(X\geq k)-nP(X\geq n+1)$$
I haven't tried showing it yet and that's because I'm confused whether it's $\sum_{k=1}^{n}P(X\geq k)-nP(X\geq n+1) $ or $\sum_{k=1}^{n}(P(X\geq k)-nP(X\geq n+1))$
so if anyone already solved this please tell me which one is the right one
Edit : really sorry I put $\infty$ instead of $n$ now it's fixed
Why do you not just solve it and find out which is the right result?
Let us start you off.
$$\sum_{k=1}^n k\, \mathsf P(X=k) ~=~ \sum_{k=1}^{n} \sum_{j=1}^k \mathsf P(X=k)$$
Because $kp=\sum\limits_{j=1}^k p$ for any positive natural number $k$.
Now, change the order of summation and continue.
PS: the result should be: $\Big(\sum_{j=1}^n \mathsf P(j\leq X)\Big) ~-~ n\,\mathsf P(n+1\leq X) $
This can be also solved with summation by parts.
With the forward difference operator $\Delta$, defined as $\Delta u_k = u_{k+1}-u_k$, you can easily derive/verify the following product rule for sequence differences (analogous to the rule for differentiation):
$$\Delta(f_k g_k) = f_k \Delta g_k + g_{n+1} \Delta f_k$$
If you sum both sides and rearrange, you get a formula for summation by parts (analogous to integration by parts):
$$\sum_{k=0}^{n} f_k \Delta g_k = f_{n+1} g_{n+1} - f_0 g_0 - \sum_{k=0}^{n} g_{k+1} \Delta f_k$$
In this case, consider $f_k = k$ and $g_k = P[X < k]$, so that $\Delta f_k = 1$ and $\Delta g_k = P[X = k]$, and you have
$$\sum_{k=0}^{n} k P[X=k] = (n+1) P[X < n+1] - 0 P[X < 0] - \sum_{k=0}^{n} P[X < k+1]$$
Finally, use $P[X < k+1] = 1 - P[X \ge k+1]$ and simplify.
