an utterly inappropriate question

perfect_radio

Let $(x_n)_{n \geq 1}$ be a sequence of real numbers s.t. $x_i > 1$ , $\forall i \in \mathbb{N}$ . Also consider $P_n=x_1 x_2 \ldots x_n$ .

Is it true that $\lim_{n \to \infty} P_n = \infty$ ?

If it turns out not to be true, then a counter-example would be appreciated Rolling Eyes

Or some sufficient conditions...

Again, sorry if this question is the dumbest ever posted here Blush

**Posted:** Thu Sep 15, 2005 2:40 pm

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Arne

We have $\lim_{n \rightarrow + \infty}\left(\prod_{i = 0}^n\left(1 + \frac{1}{2^{2^i}}\right)\right) = 2.$ So that's a nice counterexample.

Obviously we can also use the following example:

$\lim_{n \rightarrow + \infty}\left(\prod_{i = 0}^n\ \sqrt[2^i]{2}\right) = 4$ but that one is less spectacular, in my opinion Wink

Intuitively it should be obvious that the limit doesn't need to be $+\infty$ , I can easily explain why: define $y_i = \ln{x_i}$ for all $i$ . Then $\ln{P_n} = \sum_{i = 1}^ny_i$ but the only restrictions on the numbers $y_i$ is that they must be positive (since the $x_i$ are all bigger than 1). But it is obvious that an infinite sum of positive reals doesn't need to be infinite - think of geometric series with ratio smaller than 1, for example. So, take your sequence $\{y_i\}$ such that the limit $\sum_{i = 1}^{+ \infty}y_i$ is finite, then $\lim_{n \rightarrow + \infty}\ln{P_n}$ will be finite, and hence $\lim_{n \rightarrow + \infty}P_n$ will be finite too.

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perfect_radio

i thought about this while trying to solve a bloody (for me) problem (in which the product diverges).

i also tried (for that problem) taking logarithms but came up empty-handed Blush

The problem was:

Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{8k+3}{8k-3} .$

**Posted:** Mon Sep 19, 2005 12:21 am

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Arne

Well, in that case, the limit is $+ \infty$ indeed.

A proof? Well, if $a_1,a_2,...$ are positive reals then, for any $n$ , $\left(1 + a_1\right)\left(1 + a_2\right)\cdots\left(1 + a_n\right) > 1 + a_1 + a_2 + \dots + a_n.$

Now, take $a_k = \frac{6}{8k - 3}$ . Then the left hand side is exactly the given product.

Now, we get $a_1 + a_2 + \dots + a_n > \frac{6}{8} + \frac{6}{16} + \dots + \frac{6}{8n}.$

The latter sum is $\frac{3}{4}\left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right)$ and it is well known that the harmonic series diverges.

Hence, the sum must diverge as well, and hence the limit must be $+ \infty$ .

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Q-E-D

perfect_radio

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

perfect_radio

My teacher solved it in two ways:

$1^{\circ}$ Very similar to yours; it start with the same trick and then used the following generalization: $\frac{1}{a} + \frac{1}{a+r} + \frac{1}{a+2r} + \frac{1}{a+3r} + \ldots$ diverges for $a,r>0$ . The proof is a lot like yours: take $k \in \mathbb{N}$ s.t. $kr > a$ . it is clear that $\frac{1}{a+tr} > \frac{1}{a+tr+kr-a} = r \cdot \frac{1}{t+k}$ , for all $t \in \mathbb{N_0}$ . Thus, our sum is $>r \left( \frac{1}{k}+\frac{1}{k+1}+\ldots \right)$ , which famously diverges.

$2^{\circ}$ We have $\frac{8k+3}{8k-3} > \frac{8k}{8k-4} = \frac{2k}{2k-1} \Leftrightarrow -8k-12 > -24k \Leftrightarrow 16k > 12$ , which is true for all $k \geqslant 1$ . Therefore $\prod_{k=1}^{n} \frac{8k+3}{8k-3} > \prod_{k=1}^{n} \frac{2k}{2k-1}$ .
The inequality $\frac{2k}{2k-1} > \frac{2k+1}{2k} \Leftrightarrow 0 > -1$ is true for all $k \geqslant 1$ . Thus, $A = \prod_{k=1}^{n} \frac{2k}{2k-1} > \prod_{k=1}^{n} \frac{2k+1}{2k} = B$ . Multiply this inequality by $A$ to obtain $A^2 > AB = \prod_{k=1}^{n} \frac{2k(2k+1)}{2k(2k-1)} = \prod_{k=1}^{n} \frac{2k+1}{2k-1} = 2n+1$ , so $A > \sqrt{2n+1}$ . In a similar manner, $A < 2 \sqrt{n}$ .

My solution for the divergence of $A$ :

A more natural integral based approach works for estimating $A$ , but it's not that accurate.

We have $\ln A = \sum_{k=1}^n \left( \ln (2k) - \ln (2k-1) \right) = \sum_{k=1}^n \int_{2k-1}^{2k} \frac{1}{x} dx > \sum_{k=1}^n \f...$

We also have $\ln A = \sum_{k=1}^n \int_{2k-1}^{2k} \frac{1}{x} dx < \sum_{k=1}^n \frac{1}{2k-1} < 1+\ln n - \frac12 \ln(n+1) = \ln \...$ Therefore $1.334568 \cdot \sqrt{n} < A < e \cdot \sqrt{n+1}$ . Can these estimates be sharpened? (the lower bound can as seen above, but what about the upper one)

Furthermore, can this approach be used for the problem with $\frac{8k+3}{8k-3}$ to find some bounds? I tried that before seeing any solution to this problem and I didn't succeed.

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Kent Merryfield

If the terms $b_k$ are all eventually of the same sign, then
$\prod_{k=1}^{\infty}(1+b_k)$
converges if and only if
$\sum_{k=1}^{\infty}b_k$
converges.

For a proof, take the logarithm of the product.

In your case, $\frac{8k+3}{8k-3}=1+\frac6{8k-3}$ and the series diverges by comparison to the harmonic series.

**Posted:** Mon Sep 19, 2005 10:55 am

perfect_radio

I think (but can't prove) that $\left( \frac{\displaystyle \prod_{k=1}^n \frac{2k}{2k-1}}{\sqrt{n}} \right) _{n \geq 1}$ is increasing.

This implies $\lim_{n \to \infty} \frac{\displaystyle \prod_{k=1}^n \frac{2k}{2k-1}}{\sqrt{n}} = c$ , with $\sqrt{2}<c<2$ .

Can a similar approximation be found for $\prod_{k=1}^n \frac{8k+3}{8k-3}$ ?

_________________
"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Kent Merryfield

First approximation:

From $\sum_{k=1}^n\frac{c}k\approx c\ln n$ , we should get, more or less,

$\prod_{k=1}^n\left(1+\frac{c}k\right)\approx e^{c\ln n}=n^c$

(in some loose sense of approximation).

So trying to compare $\prod_{k=1}^n\frac{2k}{2k-1}$ to $\sqrt{n}$ is the direction you should be looking.

**Posted:** Mon Sep 19, 2005 4:06 pm

perfect_radio

Applying your approach to the other product would yield $\prod_{k=1}^n \frac{8k+3}{8k-3} = O \left( n^{\frac34} \right) .$ Checking with a C++ program, the constant seems to be $1.61 \ldots$ .

I'm just wondering... Is what you wrote sufficient to prove the product grows in the same manner as $n^c$ ?

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Matnomi · Hodge Conjecture Offline Joined: 16 Aug 2005 Posts: 64

**Posted:** Tue Sep 20, 2005 12:12 am

perfect_radio

**Posted:** Wed Oct 12, 2005 11:51 am

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"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."

Arne

There are two ways I know of. The first one: expand the product, you get a 1 and a sum of terms of the form $\frac{1}{2^{2^{a_1} + 2^{a_2} + ... + 2^{a_j}}}$ . Now think about binary representation of positive integers and conclude that the product is exactly $1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ... = 2$ .

Alternatively, you could use the fact that $\displaystyle 1 + \frac{1}{2^{2^i}} = \frac{\displaystyle 1 - \frac{1}{2^{2^{i + 1}}}}{\displaystyle 1 - \frac{1}{2^{2^{i}}}}$ and then you get a telescoping product.

**Posted:** Wed Oct 12, 2005 3:53 pm

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Q-E-D

perfect_radio

both solutions are ThumbUp

. thanks Arne

**Posted:** Thu Oct 13, 2005 1:04 am

_________________
"Germany has offered to send troops to the Lebanon border. I bet Israel's breathing a sigh of relief there. Nothing makes Jewish people feel safer and more secure than the German Army marching on their border."