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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.

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marked as duplicate by M Turgeon, T. Bongers, Cameron Buie, Michael Albanese, user1729 Aug 26 '13 at 19:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yeah. Couldn't find it. Can you tell me how you search for this kind of stuff? – NightRa Aug 26 '13 at 18:58
@NightRa You can try to google using your search terms and "". The google search sometimes gets you there faster. – rschwieb Aug 26 '13 at 19:54
up vote 1 down vote accepted

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).

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$$\lim_{n \to \infty} n.a^n= y$$ taking log on both sides $$\lim_{n \to \infty} log(n) + nlog(a)= log(y)$$ as $a<1$ ,$log(a) = -log(\frac{1}{a})$ and $\frac{1}{a}>1$ $$\to \lim_{n \to \infty}log(n) - nlog( \frac{1}{a} )= log(y)$$ as $ nk >log(n)$ $(k =log(\frac{1}{a})$ $$log(y) \to -\infty$$ or$$y \to 0$$

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Being just larger doesn't mean it goes to minus infinity. I had the same idea in mind. – NightRa Aug 27 '13 at 4:35

Let $\frac{1}{b} = a$ where |b|>1 therefore the limit becomes $$\lim_{n\to\infty}{n\cdot \frac{1}{b^n}}=0$$ As n increases the absolute value of the denomenator increases and therefore the absolute value of the entire expression becomes smaller and therefore converges to zero.

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(-1 for not an explanation) Saying that the absolute value of the denominator increases is correct, but why does it increase much faster than the absolute value of the numerator? Just because the denominator grows doesn't mean that the limit is zero. – user61527 Aug 26 '13 at 19:12

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