What is the $\lim_{n\rightarrow \infty }(1+\frac{1}{n})^{n^n}$

Question

What is the $$\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n^n}$$

I know that the $\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n}=e$, so I wanted to find the limit by the same way by taking $\log$ two times but I found the solution became not easy. Any help

Answer 1

Hint:Observe that: $$\left(1+\dfrac{1}{n}\right)^{n^n}> \left(\left(1+\dfrac{1}{n}\right)^n\right)^n> 2^n$$ for sufficiently large $n$.

Answer 2

Here is the solution, maybe, you are looking for:

Let $L=\lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right) ^{n^{n}},$ then \begin{eqnarray*} \log L &=&\log \lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right) ^{n^{n}} \\ &=&\lim_{n\rightarrow \infty }\log \left( 1+\frac{1}{n}\right) ^{n^{n}} \\ &=&\lim_{n\rightarrow \infty }n^{n}\log \left( 1+\frac{1}{n}\right) \\ &=&\lim_{n\rightarrow \infty }n^{n-1}\times n\log \left( 1+\frac{1}{n}% \right) \\ &=&\lim_{n\rightarrow \infty }n^{n-1}\times \frac{\log \left( 1+\frac{1}{n}% \right) }{\frac{1}{n}} \\ &=&\lim_{n\rightarrow \infty }n^{n-1}\times \lim_{n\rightarrow \infty }\frac{% \log \left( 1+\frac{1}{n}\right) }{\frac{1}{n}} \\ &=&\infty ^{\infty }\times 1=\infty . \end{eqnarray*} Then \begin{equation*} L=e^{\infty }=\infty . \end{equation*} Here I have used the following standard limit \begin{equation*} \lim_{x\rightarrow 0}\frac{\log \left( 1+x\right) }{x}=1. \end{equation*}

Answer 3

Let $$A_n=\left(1+\frac{1}{n}\right)^{n^n}$$ Taking logarithms $$\log(A_n)=n^n \log\left(1+\frac{1}{n}\right)$$ Now consider Taylor for $\log(1+x)$ when $x$ is small; in the expression, replace $x$ by $\frac 1n$ to get $$\log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ $$\log(A_n)=n^{n-1}n\Big(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+\cdots\Big)=n^{n-1}\Big(1-\frac{1}{2 n}+\frac{1}{3 n^2}+\cdots\Big)$$ So, for large $n$, $$A\approx e^{n^{n-1}}$$

Answer 4

We have, $$ \lim_{n\rightarrow \infty }n^{n-1}=\lim_{n\rightarrow \infty }\exp( \frac{n-1}{n}n\ln n) = \infty~~~\text{and}~~~\lim_{n\rightarrow \infty }\ln\left(1+\frac{1}{n}\right)^{n} =1 $$ Therefore, $$\lim_{n\rightarrow \infty }\left(1+\frac{1}{n}\right)^{n^n} =\lim_{n\rightarrow \infty }\exp\left(n^{n-1}\ln \left(1+\frac{1}{n}\right)^{n} \right) =\infty$$