# Sigma Notation and Product Notation

1. Jun 16, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
Notice that $ln [∏(k=1)^n a^k] = Ʃ_(k=1)^n * ln (a_k)$

I couldn't get the LaTeX right on this ^ But k=1 is below the product sign, and n is above. And (a^k) is the formula.
From this, as well as some calculus, calculate that:

$lim as n->∞ ∏_(k=1)^n e^\frac{k^2}{n^3}$

For this ^ the limit is as n tends to infinity, and k=1 is below the product sign, and n is above the product sign.
2. Relevant equations

3. The attempt at a solution

The first equation I think is an example of the distributive property? However, I am not sure how to show that the limit would tend to infinity for the second part, without applying actual values? (When the limit becomes 1/0 is infinity?)

Thank you.

2. Jun 16, 2013

### QED Andrew

Assume the limit exists and is equal to L. Then,

$$L = \lim_{n\to\infty} \prod_{k=1}^n e^\frac{k^2}{n^3}$$

$$\implies \ln(L) = \ln\bigl(\lim_{n\to\infty} \prod_{k=1}^n e^\frac{k^2}{n^3}\bigr)$$

$$\implies \ln(L) = \lim_{n\to\infty} \ln\bigl(\prod_{k=1}^n e^\frac{k^2}{n^3}\bigr)$$

Can you take it from here?

Last edited by a moderator: Jun 16, 2013
3. Jun 16, 2013

### Justabeginner

I think I'm oversimplifying the problem here, but would my logic work?

The index is from k=1 to n, and n goes to infinity. so lim as n-> ∞ and k=1, so k^2= 1^2= 1, and n^3= ∞, so 1/∞= 0, and e^0= 1?

Thank you.

4. Jun 16, 2013

### CAF123

You are misunderstanding the notation $\prod_{k=1}^{n}$ It means k takes on all integer values between 1 and n inclusive. So k is not always 1, it's initial value is 1.
Once you understand this, after the above posters last step, consider the hint given to you in the question.

5. Jun 17, 2013

### Justabeginner

I think the limit is infinity, after considering your and QEDAndrew's hints. Is this correct? Thank you.

6. Jun 17, 2013

### CAF123

Rather, you should say '....diverges to $+\infty$'.

I don't think it is correct. Show your work.

7. Jun 24, 2013

### Justabeginner

Okay WOW. I did not find this response in the message board, and just realized that it was in my list of unsolved problems (on here). However, I think I've got it, so I really appreciate your help CAF123. I can't believe I responded so late though. I've been working on other stuff, so pardon my absence.