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Chapter 3: Problem 14

Show that \(\lim \left((2 n)^{1 / n}\right)=1\).

Short Answer

Expert verified
The \(\lim \left((2 n)^{1 / n}\right)=1\).

Step by step solution

01

Rewrite the limit with natural logarithm

Take the natural logarithm on the both sides of the equation for the limit: \(\lim \left((2 n)^{1 / n}\right)=1\) and rewrite this equation as: \(\lim \left[\ln \left((2 n)^{1 / n}\right)\right]=\ln (1)\).
02

Apply properties of logarithm

Next, use properties of logarithm to simplify the expression: \(\lim \left[\frac{1}{n} \ln (2 n)\right] = \lim \left[\frac{\ln 2+\ln n}{n}\right]\). Using property of limit as n approach infinity, this can further simplify as \(\lim _{n \rightarrow \infty}[ \frac{\ln 2}{n} + \frac{\ln n}{n}]= 0 + \lim \left[\frac{\ln n}{n}\right]\).
03

Apply L'Hospital's rule for limit

When the function of the limit is in the form 0/0 or 鈭�/鈭�, L'Hospital's rule can be used. L'Hospital's rule states that the limit of a quotient of two functions is the limit of the quotients of their derivatives. So apply the rule and get: The limit of \(\frac{\ln n}{n}\) as n approaches infinity is equal to limit of \(\frac{1 / n}{1}\) as n approaches infinity which is equal to 0.
04

Substitute limit into original equation

Substitute the result from step 3 into the equation from step 2: \(\lim \left[\frac{\ln 2}{n} + \frac{\ln n}{n}\right]= 0 + 0 = 0\). This gives us \(\ln (1) = 0\), do the inverse operation of natural logarithm to maintain equality, we get \(e^0 = 1\). Hence, the limit of \((2n)^{1/n}\) as n approaches infinity is 1.

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Most popular questions from this chapter

If \(\sum x_{n}\) and \(\sum y_{n}\) are convergent, show that \(\sum\left(x_{n}+y_{n}\right)\) is convergent.

Establish the convergence or the divergence of the sequence \(\left(y_{n}\right)\), where $$y_{n}:=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n} \quad \text { for } \quad n \in \mathbb{N}$$

Establish the convergence and find the limits of the following sequences: (a) \(\left(\left(1+1 / n^{2}\right)^{n^{2}}\right)\). (b) \(\left((1+1 / 2 n)^{n}\right)\), (c) \(\left(\left(1+1 / n^{2}\right)^{2 n^{2}}\right)\). (d) \(\left((1+2 / n)^{n}\right)\).

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n} a_{n+1}}\) always convergent? Either prove it or give a counterexample.

Show that \(\lim \left(n^{2} / n !\right)=0\).
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