Is the sequence $f(n)$ convergent? Consider $f:\mathbb{R}\to\mathbb{R}$ is a probability density function.
(Since $‎f‎$ ‎is a‎ ‎‎probability density function then 
(1) ‎$‎‎f(x)‎\geq‎0 ‎\quad‎\text{for all} \; x$,
(2) ‎$\int_{-‎\infty‎}^{+\infty}f(x)\,dx=1$.)
Now my question is:
Let $f$ have continuous derivative on $[1,+\infty)$, Is the sequence $f(n)$ convergent? 
It seems that pdfs are convergent at infinity but I am not sure, however, I know that under some conditions such as "$f$ is uniformly continuous" we have $\lim_{x\to\infty}f(x)=0$ consequently the sequence $f(n)$ is convergent to zero. I checked the most pdf but I could not find an example of pdf wit continuous derivative such that $f(x)$ or $f(n)$ is not convergent yet. thanks.
 A: A continuous derivative is not enough to show that $f(x) \to 0$ as $x \to \infty$. $\int_{-\infty}^\infty f(x) dx = 1$ implies that the area under $f(x)$ over intervals of a given length will go to $0$ as the interval is moved to the right. But this decreasing area does not mean that $f(x)$ cannot obtain non-decreasing heights. It just means that those heights must be reached in ever-decreasing ranges. 
N.S.'s now-deleted answer showed how it can be done for continuous $f$: Have $f$ be $0$ at most locations, but around each integer $n$, have it rise up to $f(n) = n$ and drop back down fast enough that the total area contributed for $n$ is $2^{-|n|-2}$. The sum of all such contributions will be $3/4$. You can add in the missing $1/4$ in any number of ways that will not change the fact that $\lim_n f(n) = \infty$. N.S.'s example has $f(x)$ rising and falling triangularly, so it wasn't differentiable, but this is a trivial complication. One can simply use a smooth rising and falling function instead. 
(Edit changed example and added example for $f(n)$ convergent)
For example, consider the function 
$$\phi(x) =\begin{cases}0 & x\le -1\\Ae^{(x^2 - 1)^{-1}}& -1 < x< 1\\0& x \ge 1\end{cases}$$
$\phi$ is infinity differentiable at every point, including at -1 and 1, is continous and bounded above by $A$, and therefore must have a finite integral. Choose the value of $A$ so that $\int_{-\infty}^\infty \phi(x) dx = 1$. A simple substitution shows that $\int_{-\infty}^\infty \phi(ax) dx = \frac 1a$.
Now define your pdf by $$f(x) = \sum_{n-1}^\infty n\phi(n2^{n}(x-n))$$
For any given $x$, at most one term in the series is non-zero, so the series converges and $f$ is infinitely differentiable. Further
$$\int_{-\infty}^\infty f(x) dx = \sum_{n-1}^\infty n\int_{-\infty}^\infty\phi(n2^{n}(x-n))dx = \sum_{n-1}^\infty n\left(\dfrac 1{n2^n}\right) = 1$$
So $f$ is a pdf. Note that for all $n > 0$, we have $f(n) = nA$, and $f(n+1/2) = 0$. Hence $\lim_n f(n) = \infty$, while $\lim_{x\to \infty} f(x)$ has multiple limit points.
For an $f$ with $f(n)$ convergent to a non-zero value, only a minor change is needed:
$$f(x) = \sum_{n-1}^\infty \phi(2^{n}(x-n))$$
You can check that the integral of $f(x)$ is still $1$, but now $f(n) = A$ for every $n \in \Bbb N$. So $\lim_n f(x) = A$. 
$f(x)$ still diverges to multiple limit points as $x \to \infty$. But this is to be expected. While it is not true that $f(x)$ has to converge to $0$, it is true that it cannot converge to anything else. It either converges to $0$, or else it diverges to multiple limit points. To see this, note that if $f(x)$ converges to some $a > 0$, possibly even to $a = \infty$, then for any $b$ with $0 < b < a$, there is some $N$ such that if $x > N$, then $f(x) > b$. But then $$\begin{align}\int_{-\infty}^\infty f(x) dx &= \int_{-\infty}^N f(x) dx + \int_N^\infty f(x) dx \\&\ge \int_{-\infty}^N f(x) dx + \int_N^\infty b\, dx \\&= \int_{-\infty}^N f(x) dx +\infty\end{align}$$which cannot be.
