prove limit of sequence using convergence tests

I need to prove that the limit of the sequence is as shown(0):

1.$\lim_{n \to \infty} nq^n =0 ,|q|<1$

2.$\lim_{n \to \infty}\frac{2^n}{n!}$

but I need to do this using the convergence tests. With the second sequence I tried the "ratio test", and I got the result

$\lim_{n \to \infty} \frac{2}{n+1}$

which means that L in the ratio test is 0 and so it proves that the sequence converges, but how now should i prove that the limit is indeed 0? I can't use the L'Hopital's rule.

and for the first sequences I am not sure where to start.

• In your second question, I suppose your denominator is $n!$? – Idonknow Dec 12 '17 at 0:58
$$\sum\limits_{n = 1}^{ + \infty } {nq^n }$$ wit $$|q|<1$$. The series is absolutely convergent by ratio test. Namely
$$\mathop {\lim }\limits_{n \to \infty } \frac{{a_{n + 1} }} {{a_n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}} {n}\left| q \right| = \left| q \right|<1$$ Therefore the given series is also convergent. But for a convergent series it must be $$\mathop {\lim }\limits_{n \to \infty } a_n = 0$$ thus $$\mathop {\lim }\limits_{n \to \infty } nq^n = 0$$. You can follow the same way for the second exercise.
For second question, we can use the Squeeze Theorem to prove that $$\lim_{n\to\infty}\frac{2^n}{n!}=0$$ Note that for a fixed $n\in\mathbb{N},$ we have $$0 < \frac{2^n}{n!} \leq = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} ...\frac{2}{n-1}\cdot \frac{2}{n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{n-1}.$$ Since $\lim_{n\to\infty}0 = \lim_{n\to\infty}\frac{2}{n-1}=0,$ we conclude the result using Squeeze Theorem.