Pi Notation question

[DPXXPD]

Hello, I generally know of Pi Notation when it looks like this


However I have gotten a question in which it has been written like this.

            k = n-1
             ______
             |    |          k     
             |    | (1 + 1/k)  
             |    |
             k = 1

And I am unsure how exactly this is supposed to look if expanded. To be specific, it is the use of k = n-1 as the upper limit that has left me confused. Does anyone know what this expression would look like when written out and why? Thank you for your time.

 

[pka]

The expression in proper LaTeX is: \(\displaystyle \prod\limits_{k = 1}^{n - 1} {{{\left( {1 + \frac{1}{k}} \right)}^k}} \)
SEE HERE for \(\displaystyle n=11\) \(\displaystyle \prod\limits_{k = 1}^{10} {{{\left( {1 + \frac{1}{k}} \right)}^k}} = {\left( {1 + \frac{1}{1}} \right)^1} + {\left( {1 + \frac{1}{2}} \right)^2} + \cdots + {\left( {1 + \frac{1}{{10}}} \right)^{10}}\)

 

[Dr.Peterson]

I think the k on the top is just redundant, and likely a mistake. I can't think of any possible meaning for [MATH]\prod\limits_{k = 1}^{k = n - 1} {{{\left( {1 + \frac{1}{k}} \right)}^k}}[/MATH] that would make it different from [MATH]\prod\limits_{k = 1}^{n - 1} {{{\left( {1 + \frac{1}{k}} \right)}^k}}[/MATH] .

Can you show us the source, in context, so we might be able to see any subtle difference?

 

[topsquark]

The expression in proper LaTeX is: \(\displaystyle \prod\limits_{k = 1}^{n - 1} {{{\left( {1 + \frac{1}{k}} \right)}^k}} \)
SEE HERE for \(\displaystyle n=11\) \(\displaystyle \prod\limits_{k = 1}^{10} {{{\left( {1 + \frac{1}{k}} \right)}^k}} = {\left( {1 + \frac{1}{1}} \right)^1} + {\left( {1 + \frac{1}{2}} \right)^2} + \cdots + {\left( {1 + \frac{1}{{10}}} \right)^{10}}\)

Tsk tsk! Those terms are multiplied, not summed!

[math]\prod _{k = 1}^{11 - 1} \left ( 1 + \dfrac{1}{k} \right ) ^k = \left ( 1 + \dfrac{1}{1} \right ) \cdot \left ( 1 + \dfrac{1}{2} \right ) ^2 \cdot \text{ ... } \cdot \left ( 1 + \dfrac{1}{10} \right ) ^{10}[/math]
-Dan