Show that $\lim_{n \to \infty}(b_1b_2...b_n)^{\frac{1}{n}}=b$

Question

Let $b_n>0$ with the following property: $\lim_{n \to \infty}b_n=b>0$.

Show that $\lim_{n \to \infty}(b_1b_2...b_n)^{\frac{1}{n}}=b$.

I have an idea and at one moment of the proof i'm stuck. I would like to know if my idea holds and how to finish it (if it works), please.

As $(b_n)$ converges to $b$, we have by definition of convergent sequence that:

$\forall \varepsilon>0 \ \exists N>0 \ \forall n\ge N$: $|b_n-b|<\varepsilon \iff |b_n|<\varepsilon+b$.

We want to show that: $\forall \varepsilon>0 \ \exists N'>0 \ \forall n\ge N'$: $|(b_1b_2...b_n)^{\frac{1}{n}}-b|<\varepsilon \iff |b_1b_2...b_n|<(\varepsilon+b)^{n}$.

But,

$|b_1b_2...b_n|=|b_1b_2...b_N...b_n|=|b_1b_2...b_{N-1}|\cdot|b_N...b_n|<|b_1b_2...b_{N-1}|\cdot (\varepsilon+b)^{n-N+1}$

After this step i'm really stuck and don't know how to continue if it is possible. I tried as well to bound $|b_1b_2...b_{N-1}|$ by $M^{N-1}$ with $M=\max\{x_k:1\le k\le N-1\}$ but as it doesn't tend to $0$ i can't to anything using it...

P.S I know that there are some posts on this problem, but i would like to know if my approach works and if not to understand why (or have corrections). Thanks in advance for help!

Answer 1

By the AM-GM-HM inequality, for $n \in \mathbb{N} :$ $$\frac{\sum b_i}{n} \geq (b_1b_2 \dots b_n)^{\frac{1}{n}} \geq \frac{n}{\sum \frac{1}{b_i}}$$

Prove and use the following Lemma : If $\displaystyle\lim_{n \to \infty} a_n = L$ for a sequence $(a_i)_{i \in \mathbb{N}}$ where each $a_i \in \mathbb{R}$, then $\displaystyle\lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = L$.

Using the above lemma and Sandwich theorem, conclude that $(b_1b_2 \dots b_n)^{\frac{1}{n}} = b$.

Answer 2

Proceeding as you began, given $0 < \epsilon < b$, there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have

$$(b_1b_2\cdots b_N)(b -\epsilon)^{n-N+1} < (b_1b_2\cdots b_n) < (b_1b_2\cdots b_N)(b +\epsilon)^{n-N+1},$$

and, hence,

$$(b_1b_2\cdots b_N)^{\frac1{n}}(b -\epsilon)^{1-\frac{N}{n}+\frac{1}{n}} < (b_1b_2\cdots b_n)^{\frac{1}{n}} < (b_1b_2\cdots b_N)^{\frac1{n}}(b +\epsilon)^{1-\frac{N}{n}+\frac{1}{n}}$$

Holding $N$ fixed, we have

$$\lim_{n \to \infty}(b_1b_2\cdots b_N)^{\frac1{n}}(b \pm\epsilon)^{1-\frac{N}{n}+\frac{1}{n}}=1\cdot (b \pm \epsilon)$$

Thus

$$b-\epsilon <\liminf_{n \to \infty}(b_1b_2\cdots b_n)^{\frac{1}{n}} \leqslant \limsup_{n\to \infty}(b_1b_2\cdots b_n)^{\frac{1}{n}} < b + \epsilon $$

Since $\epsilon$ can be arbitrarily close to $0$, it follows that

$$b=\liminf_{n \to \infty}(b_1b_2\cdots b_n)^{\frac{1}{n}} = \limsup_{n\to \infty}(b_1b_2\cdots b_n)^{\frac{1}{n}} $$

Equality of the $\limsup$ and $\liminf$ implies that the limit exists and

$$b=\lim_{n \to \infty}(b_1b_2\cdots b_n)^{\frac{1}{n}}$$