[Math] If a is less than 1 in absolute value, than n times a to the n’th power converges to zero


Prove that if $|a|<1$ then
$$\lim_{n\to\infty}{n\cdot a^n}=0$$
Lopital's rule won't help here, and this problem appears in a book before lopital has been taught.

I noticed that:
$$\frac{a_{n+1}}{a_n}=a\frac{n+1}{n}\to a$$

So, can I conclude that it behaves like a geometric series for large n, so it converges to zero?

*Other solution methods are welcome.

Best Answer

You intuition is spot on. You have that $$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=a<1$$

Now choose a number $\ell$ with $a<\ell<1$. By the definition of limit, there must exist $N>0$ such that whenever $n\geqslant N$ we have $$\frac{a_{n+1}}{a_n}<\ell $$

In particular when $k=1,2,3,\ldots$ $$a_{N+k}<\ell a_{N+k-1}$$

By repeated use of this we can write, for $k=1,2,\ldots,$ that $$a_{N+k}<\ell ^k a_N$$

Note $N$ is fixed, so as $k\to\infty$ we get that $$\lim\limits_{k\to\infty}a_{N+k}\leqslant a_N \lim_{k\to\infty} \ell^k=0$$

Since we know $a_n\geqslant 0$; we conclude $\lim\limits_{k\to\infty} a_k=0$.

This is indeed the very idea behind D'Alambert's criterion

Let $\langle a_n\rangle $ be a sequence of positive terms, and suppose that $$\lim_{k\to\infty}\frac{a_{k+1}}{a_k}=\ell <1$$ Then $\lim\limits_{k\to\infty} a_k=0$. What's more, $\displaystyle\sum_{k\geqslant 0}a_k$ exists.

In fact, if you know about $\limsup$ and $\liminf$; we can give a finer criterion.

If $$\ell =\limsup\limits_{n\to\infty}\frac{a_{n+1}}{a_n}<1$$ then the sequences is absolutely summable. If $$\ell'=\liminf\limits_{n\to\infty}\frac{a_{n+1}}{a_n}>1$$ then the sequence is not summable (the sum diverges).