Prove that $\lim\limits_{n \to \infty} (1-\frac{1}{2n+1})^{3n} = \frac1{e\sqrt{e}}$

Question

I have this problem that I cannot seem to solve. I tried splitting it into two factors $$\lim\limits_{n \to \infty} (1-\frac{1}{2n+1})^{2n}\times \lim\limits_{n \to \infty} (1-\frac{1}{2n+1})^{n}$$ but that did not help because of the $+1$ in the denominator. So I tried multiplying both powers by $\frac{2n+1}{2n+1}$ but that did not help either. So now I'm stuck,and don't know what to do can someone please help? Thanks

Answer 1

Hint: write $3n = \dfrac{3}{2}(2n+1) - \dfrac{3}{2}$, and use the fact that $\displaystyle \lim_{n\to \infty} \left(1-\dfrac{1}{2n+1}\right)^{2n+1} = \dfrac{1}{e}$. Can you take it from here?

Answer 2

HINT: Write as $$\left(1-\frac{1}{2n+1}\right)^{3n} = \left(\left(1-\frac{1}{2n+1}\right)^{2n+1} \right)^{3/2} \left(1-\frac{1}{2n+1}\right)^{-3/2}$$

Answer 3

Let $\frac{1}{2n+1}=-t\implies t\to 0$ as $n\to \infty$, $$\lim_{n\to \infty}\left(1-\frac{1}{2n+1}\right)^{3n}$$ $$=\lim_{t\to 0}\left(1+t\right)^{-\frac{3}{2}\left(\frac 1t+1\right)}$$ $$=\lim_{t\to 0}\left(\left(1+t\right)^{\frac 1t}\right)^{-\frac 32}\cdot \lim_{t\to 0}\left(1+t\right)^{-\frac 32}$$ $$=\left(e\right)^{-\frac 32}\cdot 1$$ $$=e^{-3/2}=\color{red}{\frac{1}{e\sqrt e}}$$

Answer 4

Another way to solve the problem.

Consider $$A_n=(1-\frac{1}{2n+1})^{3n}$$ Take logarithms $$\log(A_n)=3n \log(1-\frac{1}{2n+1})$$ Now, remember that, for small $x$ $$\log(1-x)=-x-\frac{x^2}{2}+O\left(x^3\right)$$ Replace $x$ by $\frac{1}{2n+1}$ which make $$\log(A_n)=-3n\Big(\frac{1}{2n+1}+\frac 12\frac{1}{(2n+1)^2}+\cdots\Big)$$ Now, long division $$\log(A_n)=-\frac{3}{2}+\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$ $$A_n\approx e^{-3/2}(1+\frac 3 {8n})$$ which shows the limit and how it is approached.