Let $X \geq 0$ be an non-negative random variable.
Show that if $P(X > kN) \leq p^k$ for any $k \geq 1$ with some fixed $0 < p < 1$ and $N > 0$ , then $E[X] <\infty$.
My idea is to use that $E[X] = \sum_{n=0}^{\infty} P(X > n)$.
$$ E[X] = \sum_{n=0}^{\infty} P(X > n) \leq \sum_{n=0}^N P(X > n) + \sum_{n=N}^\infty P(X > N) $$
The first term is finite, but I am not sure how to argue the finiteness of the second term. I tried to start with relating the second term with the condition $P(X > kN) \leq p^k$ but didn't seem to go anywhere. i.e.
$\begin{align*} \sum_{n=N}^\infty P(X > N) = \sum_{n=N}^\infty p = \infty \\ \end{align*}$.
Any help would be greatly appreciated. Thanks in advance.
We have $$ \sum_{k=0}^{\infty} P\left(\frac XN > k\right)=\sum_{k=0}^{\infty} P\left(X >N k\right)\leqslant \sum_{k=0}^{\infty}p^k<\infty $$ hence $X/N$ has a finite expectation and so has $X$.