Mathematical induction null sequence

AI Thread Summary
The discussion focuses on proving that the sequence An = n/2^n is a null sequence, meaning it approaches zero as n approaches infinity. It references a prior proof by mathematical induction that establishes 2^n is greater than or equal to n^2 for n ≥ 5. By comparing n to 2^n, it is suggested that since 2^n grows significantly faster than n, the limit of n/2^n will indeed approach zero. The key argument is that as n increases, the denominator 2^n dominates the numerator n, leading to the conclusion that the sequence is null. Thus, it is established that the limit of An as n approaches infinity is zero.
teng125
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Show that the sequence given by An = n/2^n is a null sequence..

Hint: We have proved by mathematical induction that2^n >or equal n^2, n> or equal 5...

pls help...
 
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show that the limit n/2^n approaches 0...or that 2^n >>>> n as n goes to infinity...you know 2^n>=n^2 given above...now what can you show iwth n,n^2
 
which means n is small and 2^n is much bigger so if they are divided then the answer is appro zero??
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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