Boundedness of the sequence n^n^(-x)

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In summary, to prove that the sequence ##n^{n^{-x}}## is bounded and possibly monotonically decreasing after some point, we must use primitive tools and cannot use logarithms or the exponential constant. One approach is to look at the graph of the function ##f(n) = n^{n^{-x}}## and consider the ratio ##\frac{a_{n+1}}{a_n}## raised to the power of nx. However, it is unclear if there is enough information in the hypothesis to prove the desired result.
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bins4wins
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For any real ##x > 0##, prove that the sequence ##n^{n^{-x}}## is bounded (and if possible, monotonically decreasing after some point). The catch is that logarithms and the exponential constant cannot be used. We must arrive at the proof using fairly "primitive tools"
If you look at the graph of the function ##f(n) = n^{n^{-x}}## you'll notice that it increases at first but then decreases asymptotically towards 1 after some point. My attempt at a solution consisted of choosing three integers ##a < b < c##, then assuming that if ##a^{a^{-x}} > b^{b^{-x}}## then ##b^{b^{-x}} > c^{c^{-x}}##, but I'm not sure if there is enough information in the hypothesis to prove the desired.
 
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bins4wins said:
For any real ##x > 0##, prove that the sequence ##\large n^{n^{-x}}## is bounded (and if possible, monotonically decreasing after some point). The catch is that logarithms and the exponential constant cannot be used. We must arrive at the proof using fairly "primitive tools"

If you look at the graph of the function $f(n) = n^{n^{-x}}$ you'll notice that it increases at first but then decreases asymptotically towards 1 after some point. My attempt at a solution consisted of choosing three integers $a < b < c$, then assuming that if $a^{a^{-x}} > b^{b^{-x}}$ then $b^{b^{-x}} > c^{c^{-x}}$, but I'm not sure if there is enough information in the hypothesis to prove the desired.
Hello bins4wins. Welcome to PF !

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  • #3
Thanks, I was wondering why it wasn't working...
 
  • #4
It looks useful to consider the ratio an+1/an, raised to the power of nx.
 

FAQ: Boundedness of the sequence n^n^(-x)

What is the definition of boundedness of a sequence?

The boundedness of a sequence refers to whether the terms in the sequence are limited or restricted. A sequence is considered bounded if there is a finite number that serves as an upper or lower limit for all the terms in the sequence.

How is the sequence n^n^(-x) bounded?

The sequence n^n^(-x) is bounded because as x approaches infinity, the term n^n^(-x) approaches 0. This means that the terms in the sequence are getting progressively smaller, and there is a limit of 0 as x approaches infinity.

What is the upper bound of the sequence n^n^(-x)?

The upper bound of the sequence n^n^(-x) is 1. This can be seen by taking the limit of the sequence as x approaches infinity. As mentioned before, the terms in the sequence approach 0, and the limit of 0 is equal to 1.

Is the sequence n^n^(-x) bounded above or below?

The sequence n^n^(-x) is bounded above, as the upper bound is 1. There is no lower bound for this sequence, as the terms approach 0 but never reach it.

How does the boundedness of the sequence n^n^(-x) affect its convergence?

The boundedness of the sequence n^n^(-x) does not necessarily guarantee convergence. While the terms in the sequence are getting progressively smaller and approaching 0, there is no guarantee that the sequence will converge to a specific value. The convergence of a sequence also depends on other factors such as the rate of change of the terms and the properties of the underlying function.

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