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.

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.
