If $-1
If $-1<x<0$ then show that $nx^n \to 0$ as $n\to \infty$
How to prove this? Clearly, $x^n \to 0$ as $n\to \infty$, since $-1<x<0$. But how to prove $nx^n \to 0$ as $n\to \infty$? Is this really true?
 A: Let $(a_n)$, $(b_n)$, $(c_n)$ be the sequences defined as follows for every $n \in \mathbb{N}$: $$a_n = -n |x|^n, \qquad b_n = n x^n, \qquad c_n = n |x|^n.$$
Note: $a_n \leq b_n \leq c_n$ for every $n$. The sequence in the question is $(b_n)$. So we want to show that $\lim b_n = 0$. If we can show that $\lim a_n = \lim c_n = 0$, then $\lim b_n = 0$ by the squeeze theorem. Let's first prove that $\lim c_n = 0$.
Recall: for every sequence $(s_n)$ in $\mathbb{R}$, if $s_n > 0$ for every $n$, and if $L = \lim {(s_{n+1}/s_n)}$ exists and satisfies $L < 1$, then $\lim s_n = 0$.
Apply that result to $(c_n)$: $c_n > 0$ for every $n$, and $$\lim \dfrac{c_{n+1}}{c_n} = \lim \dfrac{n+1}{n} |x| = |x| \lim {\biggl( 1 + \dfrac{1}{n} \biggr)} = |x|(1+0) = |x| < 1.$$ Thus, $\lim c_n = 0$ by the cited result above. Then $\lim a_n = \lim -c_n = - \lim c_n = 0$. It follows that $\lim b_n = 0$ by the squeeze theorem. 
A: I assume that you know that $\sqrt[n]a\to 1$ for each $a > 0$ and even $\sqrt[n]n\to 1$.
Let $\varepsilon\in (0,1)$ and put $b_n := \sqrt[n]\varepsilon/\sqrt[n]n$.  Since $b_n\to 1$ as $n\to\infty$, there exists some $N$ such that $|x| < b_n$ for all $n\ge N$. Hence, for $n\ge N$ we have
$$
|nx^n| = n|x|^n < nb_n^n = \varepsilon.
$$
A: Inspired from Friedrich Philipp's comment, here's another proof. Note that, by the ratio test, the series
$$ \sum_{n=1}^\infty nx^n $$
converges since
\begin{align}
\lim_{n \to \infty} \left|\frac{(n+1)x^{n+1}}{nx^n} \right| &= \lim_{n \to \infty} \left|\frac{n+1}{n} \cdot x \right| \\
&= |x| \lim_{n \to \infty} \left|\frac{n+1}{n} \right| \\
&= |x| \cdot 1 \\
&= |x| < 1 \ \ \ \ \ \textrm{since $-1 < x < 0$}
\end{align}
If the series $\sum_{n=1}^\infty nx^n$ converges, then by the (contrapositive of the) $n$th Term Test, which states that

If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges

the sequence of terms $(nx^n)_{n=1}^\infty$ must tend to $0$ as $n \to \infty$.
