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Check convergence of sequences

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! :eek:

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

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


I have done the following:
  1. $\displaystyle{f_n:=\left (1-\frac{1}{2n}\right )^{3n+1}}$

    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)

  2. Could you give me a hint for that one? (Wondering)
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
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)
[/LIST]
Justify... Didn't you just do that?

That one - Does a pulsar hold still?
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
So is the first limit complete and correct? (Wondering)


About the second limit, is it as follows?

It holds that $\displaystyle{\lim_{n\rightarrow \infty}\frac{\sin n}{n}=0}$.

$(-1)^n$ is $1$ for even $n$ and $-1$ for odd $n$. So $\displaystyle{(-1)^n+\frac{\sin n}{n}}$ is $1+0=1$ for even $n$ and $-1+0=-1$ for odd $n$.

So $g_{2n} = 1$ and $g_{2n+1} = -1$.

The subsequences $g_{2n}$ and $g_{2n+1}$ have different limits ($\lim g_{2n} = 1$ and $\lim g_{2n+1} = -1$), therefore the limit $\lim g_n$ doesn't exist.


Is that correct? (Wondering)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,739
So is the first limit complete and correct?
It looks complete and correct to me. (Nod)
You have applied to multiplication and composition rules for limits, which is correct since all limits exist, and since the function in the composition is continuous. (Nerd)

About the second limit, is it as follows?

It holds that $\displaystyle{\lim_{n\rightarrow \infty}\frac{\sin n}{n}=0}$.

$(-1)^n$ is $1$ for even $n$ and $-1$ for odd $n$. So $\displaystyle{(-1)^n+\frac{\sin n}{n}}$ is $1+0=1$ for even $n$ and $-1+0=-1$ for odd $n$.

So $g_{2n} = 1$ and $g_{2n+1} = -1$.

The subsequences $g_{2n}$ and $g_{2n+1}$ have different limits ($\lim g_{2n} = 1$ and $\lim g_{2n+1} = -1$), therefore the limit $\lim g_n$ doesn't exist.


Is that correct?
I think you meant that $\lim\limits_{n\to\infty}g_{2n} = 1$ and $\lim\limits_{n\to\infty}g_{2n+1} = -1$, didn't you? Because $g_{2n} \ne 1$.
Otherwise it's all correct. (Nod)