Let $X_i$ be i.i.d random variables with common cdf F. For a given constant x, define $Y(x)= \min\{i:X_i\gt x\}$. Find $P(Y(x)\gt E[Y(x)])$. What is the limit as $x\to\infty$?
My answer:
$P(Y(x)=1)=P(X_1\gt x)=1-F$
$P(Y(x)=2)=P(X_1\lt x,X_2\gt x)=F(1-F)$
$P(Y(x)=3)=P(X_1\lt x,X_2\lt x,X_3\gt x)=F^2(1-F)$
....
$P(Y(x)=k)=F^{k-1}(1-F)$
Thus, $E[Y(x)]=\sum_{k=1}^{\infty}kF^{k-1}(1-F)={1\over 1-F}$
$P(Y(x)\gt E[Y(x)])=P(Y(x)\gt {1\over 1-F})=\sum_{k={*{1\over 1-F}*}+1}^{\infty}kF^{k-1}(1-F)$
where $*{1\over 1-F}*$ denote the largest integer less than ${1\over 1-F}$
Since $F=1$ when $x=\infty$, $E(Y(x))=\infty$, $P(Y(x)\gt E[Y(x)])=0$
Please correct me if I am wrong.
$Y(x) > E[Y(x)]$ if and only if $X_i \le x$ for all $i \le E[Y(x)]$. $$P(Y(x) > E[Y(x)]) = F(x)^{\lfloor\frac{1}{1-F(x)}\rfloor}.$$ [Your computation is close but you had a small typo. You should have $\sum_{k \ge \lfloor \frac{1}{1-F(x)}\rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x \to \infty$ leads to the indeterminate form $1^\infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as $$\frac{1}{1-F(x)} \log F(x) \le \lfloor \frac{1}{1-F(x)} \rfloor \log F(x) \le (1+\frac{1}{1-F(x)}) \log F(x).$$ You can show that the left-hand and right-hand sides tend to a common limit as $x \to \infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.