Bound on tail probability for a random variable with bounded moment and density 
Can we say that  $$\sum_{n=1}^{+\infty}\sup_{X}\mathbb P\left( |X| > n \right)<+\infty $$
where each sup is taken over any $X$ zero mean random variable with $\mathbb E[|X|]=1$, and
with distribution whose density function
is bounded by a constant $C>0$?

I tried to estimate $\mathbb P\left( |X| > n \right)$.
Markov's estimation gives $1/n$ as a bound, but it is clearly not optimal, since the bound is attained by atomic distributions, so the condition of the bound on the density is not utilized.
For example, I tried analyzing distribution with density of the kind
$$
f(x) = \begin{cases}
0 & |x|\le a\\
M|x|^{-k} & |x|>a
\end{cases}
$$
and I found that  for any $M,a,k$ satisfying the conditions, and for any $n$,
$$
\mathbb P\left( |X| > n \right)\le \frac 1e n^{-1 -\epsilon(C)}
$$
where $\epsilon = C-1 - \sqrt{C^2-2C} > 0.$
(In case you're wondering, there are no such variables with $C\le 2$)
 A: Here's a counterexample:
Let $X_n$ be a random variable with density
$$
f_n(x) = \begin{cases}
C & |x|\le a\\
C & 2n<|x|\le 2n +b\\
0 & \text{otherwise}
\end{cases}
$$
In this case we have
$$
1 = \int f_n(x) = 2C(a+b) \implies a = \frac 1{2C} -b, 0\le b<\frac 1{2C}
$$
$$
1 = \mathbb E[|X_n|] = \int |x|f_n(x) = 2C\int_0^a |x| + 2C\int_{2n}^{2n+b}|x|\\
=Ca^2 + C((2n+b)^2-(2n)^2) = C(a^2 + 4nb + b^2  )\\
=  \frac 1{4C} + (4Cn-1)b + 2Cb^2\\
\implies b = \frac 1{4C} \left[ 1-4Cn \pm \sqrt{(4Cn-1)^2 + 8C -2  }  \right]
$$
If $C>1/4$, then, for any $n$ bigger than a constant depending only on C,
$$
b = \frac {4Cn-1}{4C} \left[ -1 + \sqrt{1 + \frac{8C -2}{(4Cn-1)^2}  }  \right]\\
\ge \frac {4Cn-1}{4C} \frac{4C -1}{(4Cn-1)^2} \frac 12 = \frac{s}{4Cn-1}
$$
where $s$ is a positive constant depending only on $C$. Since
$$\sup_X\mathbb P(|X|>n) \ge \mathbb P(|X_n|>n) = bC \ge \frac{sC}{4Cn-1}$$
definitively in $n$, we conclude that the series diverges (at least as $\Omega(\log(n))$).

Update: Notice that for any $X$ with $\mathbb E[|X|]=1$, then the bound on the density cannot be less than $1/4$. If $C>1/4$, then the above counterexample works.
Surprisingly, if $C=1/4$, there's exactly one distribution for which $\mathbb E[|X|]=1$, that is
$$
f(x) = \begin{cases}
1/4 & |x|\le 2\\
0 & \text{otherwise}
\end{cases}
$$
and since it has compact support, the series converges.
