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where $n \geq 2 $

Given that a function containing an odd number of exponentiated terms as follows

$$\left(\frac1{n}\right)^{\left(\frac1{n+1}\right)^{.^{.^{.^{\left(\frac1{n+2m+1}\right)}}}}} $$

produces a decaying value

and

that a function containing an even number of exponentiated terms as follows

$$\left(\frac1{n}\right)^{\left(\frac1{n+1}\right)^{.^{.^{.^{\left(\frac1{n+2m}\right)}}}}} $$

produces a growing value

and

neither are represented by a continuous function .....

how would I prove convergence where either

a) they converge on their own values?

or

b) they converge on the same value?

or

c) they diverge completely?

In the end the main issue of concern is whether or not the following converges?

$$\lim_{m\to\infty}\left(\frac1{n}\right)^{\left(\frac1{n+1}\right)^{.^{.^{.^{\left(\frac1{m}\right)}}}}}$$

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1  
If you let $f(n)$ be the expression whose limit you're taking, $f(n)$ satisfies the functional equation $f(n+1)\ln(n)+\ln f(n)=0$. –  J. M. Apr 28 '11 at 4:42
    
@J.M. How did you get this? Fixing $n \geq 2$, if you let $f(0) = \frac{1}{n}$ and recursively define $f(m+1) = f(m)^{\frac{1}{n+m+1}}$, then $f(m)$ should satisfy the functional equation $(n+m+1)\ln(f(m+1)) - \ln(f(m)) = 0$, right? –  JavaMan Apr 28 '11 at 5:05
1  
@DJC: I was looking at the last formula, interpreting it as $\left(\frac1{n}\right)^{f(n+1)}$... –  J. M. Apr 28 '11 at 5:08
    
On a $\TeX$-nical note, I can't seem to find the "opposite" version of \ddots... –  J. M. Apr 28 '11 at 5:10
    
@J.M. I don't know why it took me so long, but now I see my mistake... –  JavaMan Apr 29 '11 at 0:21
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