I have been working on this problem for a while but couldn't find the general formula
Let
$$P(A_k) = \frac{1}{k+1}, \forall k\geq 1$$
Find:
$$P(\bigcup_{k=1}^{n}A_k)$$
Knowing that all $A_k$ mutually independent events.
I tried to use Inclusion/Exclusion formula and got this:
$$P(\bigcup_{k=1}^{n}A_k) =\sum_{k=1}^{n}\binom{n}{k}(-1)^{n-1}\frac{1}{(k+1)^{k}}$$
However, I realized that the probability changes each time so that I can't find the general formula for this. So I went on and gave it a second try but still could not derive the general form.
$$P(\bigcup_{k=1}^{n}A_k) =\sum_{k=1}^{n}\frac{1}{(k+1)}-\sum_{k=1}^{n-1}\sum_{2\leq{j}\leq{k}}^{n}\frac{(k+1)(j+1)}{(k+1)!}+...+(-1)^{n-1}\frac{1}{(k+1)!}$$
Yeah, so I'm stuck right here.
Add one $A_k$ at a time and use induction.
$P(A_1\cup...\cup A_k)=\frac{k}{k+1}$ This is true for $k=1$. Assume true for $k$.
Then $P(A_1\cup...\cup A_{k+1})=\frac{k}{k+1}+ \frac{1}{k+2}-\frac{k}{(k+1)(k+2)}=\frac{k+1}{k+2}$.