Proving the Limit of Z(n) = b for Z(n) = ((a^n + b^n))^(1/n) with 0 < a < b

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In summary, to show that the limit of Z(n) = b, given 0 < a < b, we can use the fact that Z(n) is bounded between b and 2^(1/n) * b. This means that as n approaches infinity, Z(n) will approach b. No need to use logarithms.
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semidevil
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so I need to show that Z(n) := ((a^n + b^n))^(1/n), show that the limit of Z(n) = b, given 0 < a < b.

been going back and forth and can't think of anything, but here are some random thoughts.

-basically, if I look at this in terms of definition, it is saying to show that |Z(n) - b | < episilon
- a < b means a^n < b^n
-can I conlcude that b is an upperbound?

are any of these important to solving this proof?

any ideas guys? I'm just stumped and dotn know where to start.
 
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  • #2
HINT:Use the fact that the limit and the natural logarithm commute...It should come out pretty easily.

Daniel.
 
  • #3
Forget the logarithms. You know that
Z(n) < (b^n + b^n) ^ (1/n) = 2^(1/n) * b
and
Z(n) > (b^n ) ^ (1/n) = b

Just evaluate the limits...
 
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