# Is the Convergence of These Sequences Correctly Determined?

• MHB
• mathmari
In summary, the first sequence converges to $e^{-\frac{3}{2}}$ and the second sequence does not converge as it has two sub-sequences with different limits.
mathmari
Gold Member
MHB
Hey!

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)

mathmari said:
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?

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)

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mathmari said:
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)

mathmari said:
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)

## 1. What is the definition of convergence of a sequence?

The convergence of a sequence is the concept of a sequence of numbers approaching a specific value as the number of terms in the sequence increases. In other words, as the terms in the sequence continue, the values get closer and closer to a specific number.

## 2. How do you check for convergence of a sequence?

To check for convergence of a sequence, you can use the limit comparison test, ratio test, or root test. These tests involve taking the limit of the sequence and comparing it to known values or using algebraic manipulation to determine convergence.

## 3. What is the difference between a convergent and a divergent sequence?

A convergent sequence is one in which the values approach a specific number as the number of terms increases. A divergent sequence is one in which the values do not approach a specific number and instead either increase or decrease without bound.

## 4. Can a sequence have multiple points of convergence?

No, a sequence can only have one point of convergence. If a sequence has multiple points of convergence, it is considered to be divergent.

## 5. How is the convergence of a sequence related to the convergence of a series?

The convergence of a sequence is related to the convergence of a series because a series is the sum of all the terms in a sequence. If a sequence is convergent, then the series will also be convergent. However, if a sequence is divergent, the series may still be convergent depending on the rate at which the terms in the sequence increase or decrease.

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