Sigma Notation and Product Notation

[Justabeginner]

Homework Statement

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.

Homework Equations

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.

 

[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?

 

[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.

 

[CAF123]

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.

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.

 

[Justabeginner]

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

 

[CAF123]

the limit is infinity

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

Is this correct? Thank you.

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

 

[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.