- #1

mathmari

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MHB

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Check the below sequences for convergence and determine the limit if they exist. Justify the answer.

- $\displaystyle{f_n:=\left (1-\frac{1}{2n}\right )^{3n+1}}$
- $\displaystyle{g_n:=(-1)^n+\frac{\sin n}{n}}$

- $\displaystyle{f_n:=\left (1-\frac{1}{2n}\right )^{3n+1}}$

We have that:

\begin{align*}\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{3n+1}&=\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{2n}\cdot \lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{n}\cdot \lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{1} \\ & =\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{2n}\cdot \lim_{n\rightarrow \infty}\left [\left (1-\frac{1}{2n}\right )^{2n}\right ]^{\frac{1}{2}}\cdot \lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right ) \\ & =\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{2n}\cdot \left [\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{2n}\right ]^{\frac{1}{2}}\cdot \lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right ) \\ & =\lim_{n\rightarrow \infty}\left (1+\frac{(-1)}{2n}\right )^{2n}\cdot \left [\lim_{n\rightarrow \infty}\left (1+\frac{(-1)}{2n}\right )^{2n}\right ]^{\frac{1}{2}}\cdot \lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )\end{align*}

From definition, it holds that $\displaystyle{\lim_{n\rightarrow \infty}\left (1+\frac{x}{n}\right )^n=e^x}$.

We calculate the limit $\displaystyle{\lim_{n\rightarrow \infty}\left (1+\frac{(-1)}{2n}\right )^{2n}}$ :

Let $m:=2n$. If $n\rightarrow \infty$ then $m\rightarrow \infty$.

So we get:

\begin{equation*}\lim_{n\rightarrow \infty}\left (1+\frac{(-1)}{2n}\right )^{2n}=\lim_{m\rightarrow \infty}\left (1+\frac{(-1)}{m}\right )^{m}=e^{-1}\end{equation*}

Now we consider the limit $\displaystyle{\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )}$ :

It holds the following:

\begin{equation*}\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )=\lim_{n\rightarrow \infty}1-\lim_{n\rightarrow \infty}\frac{1}{2n}=\lim_{n\rightarrow \infty}1-\frac{1}{2}\cdot \lim_{n\rightarrow \infty}\frac{1}{n}=1-\frac{1}{2}\cdot 0=1-0=1\end{equation*}

So we get:

\begin{equation*}\lim_{n\rightarrow \infty}f_n=\lim_{n\rightarrow \infty}\left (1-\frac{1}{2n}\right )^{3n+1}=e^{-1}\cdot \left [e^{-1}\right ]^{\frac{1}{2}}\cdot 1=e^{-1}\cdot e^{-\frac{1}{2}}=e^{-1-\frac{1}{2}}=e^{-\frac{3}{2} }\end{equation*} Is everything correct? So we have checked the convergence by calculating the limit, right? But how can we justify the answer? (Wondering) - Could you give me a hint for that one? (Wondering)