How to evaluate $\lim_{x \rightarrow \infty} x p^x$, where $p$ is a probability parameter? How to evaluate:  $$\lim_{x \rightarrow \infty} x p^x$$ where $p \in (0,1)$ is a probability parameter.
Should the solution be $0$, since $\lim_{x \rightarrow \infty} p^x = 0$ or should it be $\infty$, since $\lim_{x \rightarrow \infty} x = \infty$.
Edit: From here, how to evaluate: $$\lim_{x \rightarrow \infty} \sum_{n=1}^x np^n$$
I understand that $\sum_{n=0}^{\infty} np^n(1-p) = \mathbb{E}[n]=\frac {p}{1-p}$ is the expected value of the geometric distribution with parameter (1-p), Does this this help us anywhere?
 A: Consider $a=1/p$, thus $a>1$. As $a^x$ in the denominator increases at a faster rate than $x$, the limit ultimately tends to zero when $x$ becomes sufficiently large.
$$\sum\limits_{n=1}^{\infty} n/a^{n}$$ can be obtained by differentiating the geometric series of $$\sum\limits_{n=1}^{\infty} x^n$$ and putting $x=1/a$ followed by multiplication by $ 1/a$.
A: It does help, yes. Such an infinite series can only be finite if its terms approach $0$.
A: \begin{align*}
\lim_{x \rightarrow \infty} x p^x
    &= \lim_{x \rightarrow \infty} \frac{x}{p^{-x}}  \\
    &\overset{\text{l'H}}{=} \lim_{x \rightarrow \infty} \frac{1}{- \ln(p) p^{-x}}  \\
    &= \frac{-1}{\ln p} \lim_{x \rightarrow \infty} p^x  \\
    &= \frac{-1}{\ln p} \cdot 0  \\
    &= 0  \text{,}
\end{align*}
where we use l'Hospital's rule in the marked step.  We have implicitly used that $0 < p < 1$.  (If $p$ were $0$, the limit is still $0$.  If $p$ were $1$, the limit is $\infty$.)
Now to the series.  I won't solve this for you, but I will give you hints.

*

*$n p^n = (n+1)p^n - p^n$ and $\sum p^n$ is a geometric series.

*$(n+1)p^n = \frac{\mathrm{d}}{\mathrm{d}p} p^{n+1}$, then exchange the derivative and the sum, producing the derivative of a geometric series.

