#y_n=log x_n, n =2,3,4,...and y_n-(n-1)/n y_(n-1)=1/n log n#, with #y_2=log sqrt2#, how do you prove that #x_n=(n!)^(1/n)#?

Answer 1

By induction

Note that as #y_n = log(x_n)#, we have #x_n = e^(y_n)# (assuming the natural logarithm. Using a different base will not change the structure of the proof, however). Additionally, note that #y_n = 1/nlog(n)+(n-1)/ny_(n-1)# by the second given relation.

Proof: (By induction)

Base Case: For #n=2#, we are given #y_2 = log(sqrt(2))#, meaning
#x_2 = e^log(sqrt(2)) = sqrt(2) = (2!)^(1/2)#
Inductive Hypothesis: Suppose that #x_k = (k!)^(1/k)# for some integer #k>=2#.
Induction Step: We wish to show that #x_(k+1) = [(k+1)!]^(1/(k+1))#. Indeed, examining #y_(k+1)#, we have
#y_(k+1) = 1/(k+1)log(k+1)+k/(k+1)y_k#
#=1/(k+1)[log(k+1)+klog(x_k)]#
#=1/(k+1)[log(k+1)+klog((k!)^(1/k))]#
#=1/(k+1)[log(k+1)+log(k!)]#
#=1/(k+1)log(k!(k+1))#
#=1/(k+1)log((k+1)!)#
#=log([(k+1)!]^(1/(k+1)))#
meaning #x_(k+1) = e^(y_(k+1)) = [(k+1)!]^(1/(k+1))#, as desired.
We have supposed true for #k# and shown true for #k+1#, thus, by induction, #x_n = (n!)^(1/n)# for all integers #n>=2#. ∎
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Answer 2

To prove that ( x_n = (n!)^{1/n} ) given the recursive relation ( y_n = \log{x_n} ) for ( n = 2, 3, 4, \ldots ) and ( y_n - \frac{n-1}{n} y_{n-1} = \frac{1}{n} \log{n} ), with ( y_2 = \log{\sqrt{2}} ), we can proceed as follows:

  1. We start by rewriting the given recursive relation: [ y_n - \frac{n-1}{n} y_{n-1} = \frac{1}{n} \log{n} ]

  2. Since ( y_n = \log{x_n} ), we substitute this into the equation: [ \log{x_n} - \frac{n-1}{n} \log{x_{n-1}} = \frac{1}{n} \log{n} ]

  3. Taking exponentials of both sides, we have: [ x_n = x_{n-1}^{\frac{n-1}{n}} \cdot n^{\frac{1}{n}} ]

  4. Now, we apply the initial condition ( y_2 = \log{\sqrt{2}} ), which gives us ( x_2 = \sqrt{2} ).

  5. We can now use the recursive relation to find subsequent values of ( x_n ): [ x_3 = x_2^{\frac{2}{3}} \cdot 3^{\frac{1}{3}} ] [ x_4 = x_3^{\frac{3}{4}} \cdot 4^{\frac{1}{4}} ] [ \vdots ] [ x_n = x_{n-1}^{\frac{n-1}{n}} \cdot n^{\frac{1}{n}} ]

  6. By induction, we can show that for any ( n ), ( x_n = (n!)^{1/n} ) holds.

Thus, by recursively applying the given relation and using the initial condition, we can prove that ( x_n = (n!)^{1/n} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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