Recurence

[markraz]

Hi I have this explicit formula and need to turn it into a Recurrence relation

Explicit formula: \(\displaystyle ( \frac{1}{k})^{k} \)

Question.... How do you get from the "Explicit formula" to the "Recurrence Relation" mathematically/systematically?

So far I set it up as follows:

\(\displaystyle a_{k}=( \frac{1}{k})^{k}
\)

\(\displaystyle
a_{k+1}=( \frac{1}{(k+1)})^{(k+1)}
\) (not even sure if this is setup right??)

Anyone know what's next? to get to Recurrence Relation:??


\(\displaystyle
a_{n}=??
\)

Thanks in advance

 

[markraz]

pretty please?

 

[Ishuda]

pretty please?

I haven't been able to find a nice relationship but just for grins and giggles:
\(\displaystyle \frac{a_{k+1}}{a_k} = \frac{1}{k+1}\frac{1}{(\frac{k+1}{k})^k}= \frac{1}{k+1}\frac{1}{(1 + \frac{1}{k})^k} \to \frac{1}{k+1}\frac{1}{e} = e^{-1} \frac{1}{k+1}\)
or
\(\displaystyle a_{k+1} \to e^{-1} \frac{1}{k+1} = e^{-1} \frac{a_k}{1 + a_k} \)

or, another
\(\displaystyle a_{k+1} = (\frac{a_k}{1 + a_k})^{k+1} = (\frac{a_k}{1 + a_k})^{a_k^{-1}+1}\)

 

[markraz]

I haven't been able to find a nice relationship but just for grins and giggles:
\(\displaystyle \frac{a_{k+1}}{a_k} = \frac{1}{k+1}\frac{1}{(\frac{k+1}{k})^k}= \frac{1}{k+1}\frac{1}{(1 + \frac{1}{k})^k} \to \frac{1}{k+1}\frac{1}{e} = e^{-1} \frac{1}{k+1}\)
or
\(\displaystyle a_{k+1} \to e^{-1} \frac{1}{k+1} = e^{-1} \frac{a_k}{1 + a_k} \)

or, another
\(\displaystyle a_{k+1} = (\frac{a_k}{1 + a_k})^{k+1} = (\frac{a_k}{1 + a_k})^{a_k^{-1}+1}\)

Thank you