How do I $n\lambda ^n$ tends to $0$? I'm working with sequences of functions, and I have a question about a limit:
If $$0<\lambda <1$$
then, when $n \rightarrow \infty$, $n$ natural number,
$$(n\lambda ^n)_{n\in \mathbb{N}} \rightarrow 0\, \, ?$$ 
How can I prove it? Thanks for any hint!
 A: Using infinite series:
$$a_n:=n\lambda^n\implies \frac{a_{n+1}}{a_n}=\frac{n+1}n\lambda\xrightarrow[n\to\infty]{}\lambda<1$$
and thus by the ratio test (D'Alembert's Test) , the series
$$\sum_{n=1}^\infty n\lambda^n\;\;\text{converges}\;\implies\;n\lambda^n\xrightarrow[n\to\infty]{}0$$
A: A plain old pre-calculus series and sequences demonstration:
For $0 < \lambda < 1$, we have $1 < \lambda^{-1}$ whence $0 < \lambda^{-1} - 1$.  Thus there exists a positive $N$ such that $0 < N^{-1} < \lambda^{-1} - 1$; then $1 + N^{-1} < \lambda^{-1}$; multiplying this inequality through by $N\lambda$ yields $(N + 1)\lambda < N$ or $((N + 1)/N)\lambda < 1$; this last may be written $(1 + (1/N))\lambda < 1$.  Taking integer $n \ge N$ shows that $(1 + (1/n))\lambda \le (1 + (1/N)) \lambda < 1$ or $((n + 1)/n)\lambda \le (1 + (1/N)) \lambda < 1$ for all such $n$.  Thus there exists $\rho$, $0 < \rho < 1$, with $(1 + (1/N))\lambda < \rho < 1$ and
$\dfrac{(n + 1) \lambda^{n + 1}}{n \lambda^n} = \dfrac{(n + 1) \lambda}{n} \dfrac{\lambda^n}{\lambda^n} = \dfrac{(n + 1)\lambda}{n} \le (1 + (1/N))\lambda < \rho < 1, \tag{1}$
under the proviso that $n \ge N$.  Thus,
$(n + 1)\lambda^{n + 1} < \rho (n \lambda^n) \tag{2}$
for all $n \ge N$ sufficiently large.  But then
$(n + 2)\lambda^{n + 2} < \rho ((n + 1) \lambda^{n + 1})  < \rho^2 (n \lambda^n), \tag{3}$
and it is pretty easy to see that continuing in this manner leads to
$(n + m)\lambda^{n + m} < \rho^m (n\lambda^n) \to 0 \; \text{as} \; m \to \infty. \tag{4}$
(2) and (4) show that once the sequence $n \lambda^n$ progresses to the point where $n \ge N$, it decreases monotonically to zero as $n \to \infty$.  QED.
And that's how it may be shown!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Instead of $$\frac{\lambda^x}{1/x}$$
use $$\frac{x}{\lambda^{-x}}$$
in L'Hopital's rule.
