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Finding the lim inf and lim sup of a sequence

Prove that a sequence converges to a finite limit iff lim inf equals lim supLim Sup and Lim InfShow that $lim inf a_nleliminf s_n.$Set lim inf and lim sup question.Finding lim sup and lim infIf $x_n$ Cauchy then $limsup x_n = liminf x_n$Unsure if this proof works for proving $supgeqinf$Show that $liminf A_nsubset limsup_n A_n$lim inf and lim sup of sequence obtained from combination of two sequencesSequence $a_n$ such that $inf a_n < lim inf a_n < lim sup a_n < sup a_n$

2

1

$begingroup$

I'm working on the following problem:

Find $lim sup_n a_n$ and $liminf_n a_n$ of the sequence given by $a_n = 1 + (-1)^nfrac{2n+3}n.$

I've done the following work so far:

Let ${i_n}_n$ be the sequence given by $i_n = inf{a_k:kge n}.$ Then, $i_n = {-4, -2, frac {-8}5, frac{-10}7,cdots}.$

I know that the $lim inf_na_n = lim_n i_n$, but I can't figure out how to take the limit of $i_n$.

real-analysis sequences-and-series limsup-and-liminf

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asked 15 hours ago

B RetnikB Retnik

534

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2

1

$begingroup$

I'm working on the following problem:

Find $lim sup_n a_n$ and $liminf_n a_n$ of the sequence given by $a_n = 1 + (-1)^nfrac{2n+3}n.$

I've done the following work so far:

Let ${i_n}_n$ be the sequence given by $i_n = inf{a_k:kge n}.$ Then, $i_n = {-4, -2, frac {-8}5, frac{-10}7,cdots}.$

I know that the $lim inf_na_n = lim_n i_n$, but I can't figure out how to take the limit of $i_n$.

real-analysis sequences-and-series limsup-and-liminf

share|cite|improve this question

asked 15 hours ago

B RetnikB Retnik

534

$endgroup$

add a comment | 

$begingroup$

I'm working on the following problem:

Find $lim sup_n a_n$ and $liminf_n a_n$ of the sequence given by $a_n = 1 + (-1)^nfrac{2n+3}n.$

I've done the following work so far:

Let ${i_n}_n$ be the sequence given by $i_n = inf{a_k:kge n}.$ Then, $i_n = {-4, -2, frac {-8}5, frac{-10}7,cdots}.$

I know that the $lim inf_na_n = lim_n i_n$, but I can't figure out how to take the limit of $i_n$.

real-analysis sequences-and-series limsup-and-liminf

share|cite|improve this question

asked 15 hours ago

B RetnikB Retnik

534

$endgroup$

share|cite|improve this question

asked 15 hours ago

B RetnikB Retnik

534

share|cite|improve this question

asked 15 hours ago

B RetnikB Retnik

534

asked 15 hours ago

B RetnikB Retnik

534

asked 15 hours ago

B RetnikB Retnik

534

2 Answers
2

active

oldest

votes

3

$begingroup$

Let's first notice that $frac{2n+3}{n}$ converges to $2$. Therefore, the sequence
$$(-1)^nfrac{2n+3}{n}$$

has only two limit points, which are $2$ and $-2$. This implies that $(a_n)$ has only two limit points, which are $3$ and $-1$. Therefore, you get
$$liminf a_n = -1 quad text{and} quad limsup a_n = 3$$

share|cite|improve this answer

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

1

$begingroup$

Let's compute the even and odd terms. So, as $nrightarrow infty $, we have
begin{eqnarray*}
a_{2n} &=&1+frac{4n+3}{2n}=frac{3n+3/2}{n}to3, \
a_{2n-1} &=&1-frac{2left( 2n-1right) +3}{2n-1}=-frac{n+1}{n-1/2}to-1.
end{eqnarray*}

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

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add a comment | 

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3

$begingroup$

Let's first notice that $frac{2n+3}{n}$ converges to $2$. Therefore, the sequence
$$(-1)^nfrac{2n+3}{n}$$

has only two limit points, which are $2$ and $-2$. This implies that $(a_n)$ has only two limit points, which are $3$ and $-1$. Therefore, you get
$$liminf a_n = -1 quad text{and} quad limsup a_n = 3$$

share|cite|improve this answer

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

3

$begingroup$

Let's first notice that $frac{2n+3}{n}$ converges to $2$. Therefore, the sequence
$$(-1)^nfrac{2n+3}{n}$$

has only two limit points, which are $2$ and $-2$. This implies that $(a_n)$ has only two limit points, which are $3$ and $-1$. Therefore, you get
$$liminf a_n = -1 quad text{and} quad limsup a_n = 3$$

share|cite|improve this answer

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

add a comment | 

$begingroup$

Let's first notice that $frac{2n+3}{n}$ converges to $2$. Therefore, the sequence
$$(-1)^nfrac{2n+3}{n}$$

has only two limit points, which are $2$ and $-2$. This implies that $(a_n)$ has only two limit points, which are $3$ and $-1$. Therefore, you get
$$liminf a_n = -1 quad text{and} quad limsup a_n = 3$$

share|cite|improve this answer

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

$endgroup$

share|cite|improve this answer

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

answered 15 hours ago

TheSilverDoeTheSilverDoe

3,011112

1

$begingroup$

Let's compute the even and odd terms. So, as $nrightarrow infty $, we have
begin{eqnarray*}
a_{2n} &=&1+frac{4n+3}{2n}=frac{3n+3/2}{n}to3, \
a_{2n-1} &=&1-frac{2left( 2n-1right) +3}{2n-1}=-frac{n+1}{n-1/2}to-1.
end{eqnarray*}

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

$endgroup$

add a comment | 

1

$begingroup$

Let's compute the even and odd terms. So, as $nrightarrow infty $, we have
begin{eqnarray*}
a_{2n} &=&1+frac{4n+3}{2n}=frac{3n+3/2}{n}to3, \
a_{2n-1} &=&1-frac{2left( 2n-1right) +3}{2n-1}=-frac{n+1}{n-1/2}to-1.
end{eqnarray*}

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

$endgroup$

add a comment | 

$begingroup$

Let's compute the even and odd terms. So, as $nrightarrow infty $, we have
begin{eqnarray*}
a_{2n} &=&1+frac{4n+3}{2n}=frac{3n+3/2}{n}to3, \
a_{2n-1} &=&1-frac{2left( 2n-1right) +3}{2n-1}=-frac{n+1}{n-1/2}to-1.
end{eqnarray*}

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206

answered 13 hours ago

Américo TavaresAmérico Tavares

32.5k1180206