Prove that $\lim_{n\to \infty} n\cdot r^n=0$ where $(0\leq r <1)$ without using ratio test $\lim_{n\to \infty} n\cdot r^n=0$, where $0\leq r <1$, can be obtained by vanishing condition (considering $\sum^{\infty}_{n=1}n\cdot r^n$, which converges, using ratio test).
Is there a direct way of finding this limit without using ratio test? It seems quite easy but I haven't come up with an idea yet.
 A: Write $\frac 1r = a + 1$ where $a > 0$. If  $n \ge 2$ you have $$\frac 1{r^n} = (a + 1)^n \ge \frac{n(n-1)}{2} a^2$$ according to the binomial theorem. Consequently
$$0 \le n r^n \le \frac{2}{(n-1)a^2}$$ for all $n \ge 2$. Now let $n \to \infty$ and use the squeeze theorem.
A: The case $r=0$ is quite trivial, hence we assume $r\in(0,1)$. Let $s=-\log r\in\mathbb{R}^+$.
We have to prove:
$$ \lim_{n\to +\infty} n\cdot e^{-sn}=0, \tag{1}$$
that follows from:
$$ 0\leq n\cdot e^{-sn} = \frac{n}{\left(e^{\frac{s}{2}n}\right)^2}\leq\frac{n}{\left(1+\frac{s}{2}n\right)^2}\leq\frac{4}{s^2 n}.\tag{2}$$
A: You can use $  Abel-Pringsheim $  $Theorem $ Here
Abel's (or Pringsheim's) theorem: If $ \sum U_n$ is a convergent series of positive and decreasing terms, then $lim $ $ nU_n = 0$.
Consider the the series $\sum_{k=1}^{n} r^n $ where $0 \leq r<1$ 
apply $  Abel-Pringsheim$ $ Theorem $
A: We can prove this by contradiction. Suppose that the sequence does not go to zero, then there is an $\epsilon >0$ for which given any $N \in \mathbb{N}$ there is an integer $n > N$ for which $n r^n > \epsilon$. Therefore we have $$r > \epsilon^{1/n} \left( \frac1n \right)^{1/n}.$$
Since we can make $n$ as large as we like, we can then take the limit of the right hand side to determine a lower bound for $r$. (This is because the function $\epsilon^{1/x} (1/x)^{1/x}$ is increasing for large enough $x$.)
We know that $\epsilon^{1/n} \to 1$, and $(1/n)^{1/n}\to 1$ can be seen as well after an application of the logarithm and using l'Hopital's rule.
Thus $r \ge 1$, which is a contradiction.
