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Let $b∈\mathbb{R}$ satisfy $0<b<1$. Show that $\lim_\limits{n \to \infty}(nb^n)=0$

How do I solve this problem?

Student
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vishu
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    And the limit is for $n\to\infty$ or $n\to 0$? Cause for $n\to 0$ it is trivial ;-) – Maczinga Mar 04 '17 at 18:30
  • @Maczinga n tends to infinity – vishu Mar 04 '17 at 18:35
  • Also: http://math.stackexchange.com/questions/1571231/find-limit-for-ncn-when-c1, http://math.stackexchange.com/questions/1132422/find-the-limit-of-function-f-nx-nxn. – Martin R Mar 04 '17 at 18:48

2 Answers2

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Method 1

Put $u_n=nb^n$ and take its logarithm :

$$\ln(u_n)=\ln(n)+n\ln(b)=n\left(\frac{\ln(n)}{n}+\ln(b)\right)$$Since we know that $\lim_{n\to\infty}\frac{\ln(n)}{n}=0$, we conclude that $\lim_{n\to\infty}\ln(u_n)=-\infty$ and hence that $\lim_{n\to\infty}u_n=0$

Method 2

Put again $u_n=nb^n$. We see that :

$$\frac{u_{n+1}}{u_n}=\frac{n+1}{n}b\underset{n\to\infty}{\longrightarrow}b<1$$

which implies that the sequence $(nb^n)_{n\ge N}$ is decreasing (for some convenient $N\in\mathbb{N}^\star$).

Being bounded from below, it converges to some limit $L\ge0$. But, taking the limits of both sides of $u_{n+1}=\frac{n+1}{n}bu_n$, we get $L=bL$, hence $L=0$.

Method 3 (the shortest)

By the ratio test, the series $\sum_{n\ge1}nb^n$ is convergent, and hence $\lim_{n\to\infty}nb^n=0$

Adren
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If $0<b<1$ then there exists $c\in\Bbb{R}$ such that $b=\frac1{c}$ and $c>1$. Hence the limit becomes:

$\lim_{n\to\infty}\frac{n}{c^n}$ which is well-known to converge to $0$.

Maczinga
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