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Show that $\lim\limits_{n\rightarrow\infty}(nb^n)=0$ for $0<b<1$.

I have seen a couple different proofs of this using logarithms, the Binomial Theorem, and Bernoulli's Inequality. Although there are duplicates of this question on this site, I'd like to check my method of proof for rigor:

Proof:

Since $b\in(0,1)$, we can write $\frac{1}{\sqrt{b}}=1+d$ for some $d>0$. By Bernoulli's Inequality, we know $(1+d)^n>1+nd\space\space\forall n\in\mathbb{N}$. Rearranging our first equation for $b$ we get: $$\frac{1}{b}=(1+d)^2\implies b^n=\frac{1}{((1+d)^2)^n}\implies nb_n=\frac{n}{((1+d)^2)^n}$$

Showing $\lim\limits_{n\rightarrow\infty}(nb^n)=0,$ $$|nb^n-0|=nb^n=\frac{n}{((1+d)^2)^n}<\frac{n}{(1+nd)^2}=\frac{n}{n^2d^2+2nd+1}<\bigg(\frac{1}{d^2}\bigg)\bigg(\frac{1}{n}\bigg)$$

Now I will apply the theorem that if $(a_n)$ is a sequence of positive real numbers with $\lim\limits_{n\rightarrow\infty}(a_n)=0$ and $C>0$ and $|x_n-x|\leq Ca_n$ , then $\lim\limits_{n\rightarrow\infty}x_n=x.$ So taking $C=\frac{1}{d^2}$ since $d>0$ and $(a_n)=\frac{1}{n}$, we have that $$|nb^n-0|<\bigg(\frac{1}{d^2}\bigg)\bigg(\frac{1}{n}\bigg)$$

Conclude that indeed $\lim\limits_{n\rightarrow\infty}(nb^n)=0$. $\blacksquare$

Thanks in advance for the help!

rtybase
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coreyman317
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  • Note that your $d_n$ is a constant by definition. It is not a good idea to use the notation suggesting it depends on $n$. – dan_fulea Nov 07 '18 at 12:24
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    Related https://math.stackexchange.com/questions/2976535/how-to-prove-that-x-n-nqn-for-q-1-is-bounded ... it contains a proof using binomial theorem – rtybase Nov 07 '18 at 12:38
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    Not an answer, just a quick proof of the same result: if $x_n = nb^n$, then for all $n > b/(1 - b)$ we have $x_n/x_{n+1} = n/((n+1)b) > 1$, so $(x_n)$ is eventually decreasing and bounded below by $0$, therefore it tends to some $l \geqslant 0$, and $l = \lim_{n\to\infty}(nb^n) = \lim_{n\to\infty}((n+1)b^{n+1}) = \lim_{n\to\infty}((1 + 1/n)b(nb^n)) = bl$, i.e. $(1 - b)l = 0$, i.e. $l = 0$. – Calum Gilhooley Nov 07 '18 at 18:38
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    This proof is even quicker, and has even fewer prerequisites, but it is arguably a bit strained: for all $n \geqslant 2b/(1-b)$, we have $x_n/x_{n+1} \geqslant 2/(1+b) > 1$; therefore, $x_n \to l \geqslant 0$, just as before; and now, we have $(1+b)l = \lim_{n\to\infty}((1+b)x_n) \geqslant \lim_{n\to\infty}(2x_{n+1}) = 2l$, i.e. $(1-b)l \leqslant 0$, therefore $l \leqslant 0$, therefore $l = 0$. – Calum Gilhooley Nov 07 '18 at 20:36

1 Answers1

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Yor proof is correct. But why do you write $d_n$ ? In $\frac{1}{\sqrt{b}}=1+d_n$ the number $d_n $ is independent of $n$.

A simpler proof: by the root test the series $ \sum nb^n$ is convergent, hence $nb^n \to 0$.

Fred
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