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heshbon
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Homework Statement
Need to prove n^(1/n) tend to 1 as n tends to infinty
Homework Equations
The Attempt at a Solution
Have tried comparing to n^(1/n)=(1+h) and using binomial series but no joy..please help
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The sequence [itex]n^{1/n}[/itex], as n goes to infinity, converges to a if the function [itex]x^{1/x}[/itex] converges to a as x goes to infinity. If we set [itex]y= x^{1/x}[/itex] then ln(y)= (ln x)/x which is of the "infinity/infinity" form so we can use L'Hopital's rule.heshbon said:Homework Statement
Need to prove n^(1/n) tend to 1 as n tends to infinty
Homework Equations
The Attempt at a Solution
Have tried comparing to n^(1/n)=(1+h) and using binomial series but no joy..please help
I should have included the next term in the binomial expansionPere Callahan said:The first equation is equivalent to [itex]n\leq(1+\varepsilon)^n=1+n\varepsilon+\dots[/itex]
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The expression n^(1/n) is known as the n-th root of n. It is the number that, when multiplied by itself n times, gives the value of n.
Proving that n^(1/n) tends to 1 as n tends to infinity is important because it helps us understand the behavior of the function as the input (n) grows larger and larger. It also has applications in fields such as calculus, number theory, and statistics.
The expression n^(1/n) can be simplified as n^(1/n) = (n^1)^(1/n) = n^(1/n) = n^(1/n) = n^(1/n) = n^(1/n) = n^(1/n).
The proof involves using the binomial theorem and the concept of limits. It can be shown that as n increases, the terms involving n in the binomial theorem become smaller and smaller, ultimately approaching 0. This leads to the limit of n^(1/n) approaching 1 as n tends to infinity.
Yes, there are exceptions to this result. For example, when n is a negative number or a fraction, the expression n^(1/n) does not tend to 1 as n tends to infinity. It is also important to note that this result is only true for real numbers, not complex numbers.