Proof verification of $x_n = frac{n^2 - 2n}{n+1}$ diverges using $varepsilon$ definition.
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Given
$$
x_n = frac{n^2 - 2n}{n+1}
$$
Prove that ${x_n}$ diverges.
I've started as follows. Suppose there sequence converges to some value. That means:
$$
lim_{n to infty}frac{n^2-2n}{n+1} = L
$$
Consider $x_n$:
$$
frac{n^2-2n}{n+1} = frac{n(n-2)}{n+1} < frac{n^2}{n+1} < frac{n^2}{n} le n
$$
Now by definition of the limit we want:
$$
|x_n - L| =left|frac{n^2 - 2n}{n+1} - Lright| <|n-L| < varepsilon
$$
Suppose $varepsilon = 1$, then:
$$
|n-L| < 1
$$
But since $L$ is fixed we may by Archimedean principle find $n$ such that $n > L implies n + 1 > L$ and therefore the limit doesn't exist.
I'm kindly asking to verify my proof or correct me if it's wrong.
calculus limits proof-verification epsilon-delta
asked Nov 23 at 13:57
roman
1,50111119
add a comment |
up vote
0
down vote
favorite
Given
$$
x_n = frac{n^2 - 2n}{n+1}
$$
Prove that ${x_n}$ diverges.
I've started as follows. Suppose there sequence converges to some value. That means:
$$
lim_{n to infty}frac{n^2-2n}{n+1} = L
$$
Consider $x_n$:
$$
frac{n^2-2n}{n+1} = frac{n(n-2)}{n+1} < frac{n^2}{n+1} < frac{n^2}{n} le n
$$
Now by definition of the limit we want:
$$
|x_n - L| =left|frac{n^2 - 2n}{n+1} - Lright| <|n-L| < varepsilon
$$
Suppose $varepsilon = 1$, then:
$$
|n-L| < 1
$$
But since $L$ is fixed we may by Archimedean principle find $n$ such that $n > L implies n + 1 > L$ and therefore the limit doesn't exist.
I'm kindly asking to verify my proof or correct me if it's wrong.
calculus limits proof-verification epsilon-delta
asked Nov 23 at 13:57
roman
1,50111119
1
You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given
$$
x_n = frac{n^2 - 2n}{n+1}
$$
Prove that ${x_n}$ diverges.
I've started as follows. Suppose there sequence converges to some value. That means:
$$
lim_{n to infty}frac{n^2-2n}{n+1} = L
$$
Consider $x_n$:
$$
frac{n^2-2n}{n+1} = frac{n(n-2)}{n+1} < frac{n^2}{n+1} < frac{n^2}{n} le n
$$
Now by definition of the limit we want:
$$
|x_n - L| =left|frac{n^2 - 2n}{n+1} - Lright| <|n-L| < varepsilon
$$
Suppose $varepsilon = 1$, then:
$$
|n-L| < 1
$$
But since $L$ is fixed we may by Archimedean principle find $n$ such that $n > L implies n + 1 > L$ and therefore the limit doesn't exist.
I'm kindly asking to verify my proof or correct me if it's wrong.
calculus limits proof-verification epsilon-delta
asked Nov 23 at 13:57
roman
1,50111119
Given
$$
x_n = frac{n^2 - 2n}{n+1}
$$
Prove that ${x_n}$ diverges.
I've started as follows. Suppose there sequence converges to some value. That means:
$$
lim_{n to infty}frac{n^2-2n}{n+1} = L
$$
Consider $x_n$:
$$
frac{n^2-2n}{n+1} = frac{n(n-2)}{n+1} < frac{n^2}{n+1} < frac{n^2}{n} le n
$$
Now by definition of the limit we want:
$$
|x_n - L| =left|frac{n^2 - 2n}{n+1} - Lright| <|n-L| < varepsilon
$$
Suppose $varepsilon = 1$, then:
$$
|n-L| < 1
$$
But since $L$ is fixed we may by Archimedean principle find $n$ such that $n > L implies n + 1 > L$ and therefore the limit doesn't exist.
I'm kindly asking to verify my proof or correct me if it's wrong.
calculus limits proof-verification epsilon-delta
calculus limits proof-verification epsilon-delta
asked Nov 23 at 13:57
roman
1,50111119
asked Nov 23 at 13:57
roman
1,50111119
asked Nov 23 at 13:57
roman
1,50111119
asked Nov 23 at 13:57
roman
1,50111119
asked Nov 23 at 13:57
roman
1,50111119
1,50111119
1
You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16
add a comment |
1
You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16
1
1
You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16
You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
You can do it like this: Fix an $L>0.$ $x_n>LRightarrow n^2-2n>nL+LRightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 Rightarrow n>L/2+1+sqrt{L+(L/2+1)^2} .$
So, for an arbitrary $L$ if we choose $n_0=[L/2+1+sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $ngeq n_0$. Hence $x_nrightarrow infty.$
answered Nov 23 at 14:09
John_Wick
1,224111
add a comment |
up vote
1
down vote
You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.
To prove that it diverges, recall that by the definition of divergent limit, for any $Min mathbb{R}$ we need to show that eventually
$$frac{n^2 - 2n}{n+1} ge M iff n^2-(2+M)n-M ge 0$$
answered Nov 23 at 14:06
gimusi
93k94495
add a comment |
up vote
1
down vote
The error lies in this step:
$|frac{n^2 - 2n}{n+1} - L| <|n-L|$
You know that $frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?
answered Nov 23 at 14:14
Shubham Johri
2,819413
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can do it like this: Fix an $L>0.$ $x_n>LRightarrow n^2-2n>nL+LRightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 Rightarrow n>L/2+1+sqrt{L+(L/2+1)^2} .$
So, for an arbitrary $L$ if we choose $n_0=[L/2+1+sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $ngeq n_0$. Hence $x_nrightarrow infty.$
answered Nov 23 at 14:09
John_Wick
1,224111
add a comment |
up vote
1
down vote
accepted
You can do it like this: Fix an $L>0.$ $x_n>LRightarrow n^2-2n>nL+LRightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 Rightarrow n>L/2+1+sqrt{L+(L/2+1)^2} .$
So, for an arbitrary $L$ if we choose $n_0=[L/2+1+sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $ngeq n_0$. Hence $x_nrightarrow infty.$
answered Nov 23 at 14:09
John_Wick
1,224111
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can do it like this: Fix an $L>0.$ $x_n>LRightarrow n^2-2n>nL+LRightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 Rightarrow n>L/2+1+sqrt{L+(L/2+1)^2} .$
So, for an arbitrary $L$ if we choose $n_0=[L/2+1+sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $ngeq n_0$. Hence $x_nrightarrow infty.$
answered Nov 23 at 14:09
John_Wick
1,224111
You can do it like this: Fix an $L>0.$ $x_n>LRightarrow n^2-2n>nL+LRightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 Rightarrow n>L/2+1+sqrt{L+(L/2+1)^2} .$
So, for an arbitrary $L$ if we choose $n_0=[L/2+1+sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $ngeq n_0$. Hence $x_nrightarrow infty.$
answered Nov 23 at 14:09
John_Wick
1,224111
answered Nov 23 at 14:09
John_Wick
1,224111
answered Nov 23 at 14:09
John_Wick
1,224111
answered Nov 23 at 14:09
John_Wick
1,224111
1,224111
add a comment |
add a comment |
up vote
1
down vote
You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.
To prove that it diverges, recall that by the definition of divergent limit, for any $Min mathbb{R}$ we need to show that eventually
$$frac{n^2 - 2n}{n+1} ge M iff n^2-(2+M)n-M ge 0$$
answered Nov 23 at 14:06
gimusi
93k94495
add a comment |
up vote
1
down vote
You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.
To prove that it diverges, recall that by the definition of divergent limit, for any $Min mathbb{R}$ we need to show that eventually
$$frac{n^2 - 2n}{n+1} ge M iff n^2-(2+M)n-M ge 0$$
answered Nov 23 at 14:06
gimusi
93k94495
add a comment |
up vote
1
down vote
up vote
1
down vote
You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.
To prove that it diverges, recall that by the definition of divergent limit, for any $Min mathbb{R}$ we need to show that eventually
$$frac{n^2 - 2n}{n+1} ge M iff n^2-(2+M)n-M ge 0$$
answered Nov 23 at 14:06
gimusi
93k94495
You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.
To prove that it diverges, recall that by the definition of divergent limit, for any $Min mathbb{R}$ we need to show that eventually
$$frac{n^2 - 2n}{n+1} ge M iff n^2-(2+M)n-M ge 0$$
answered Nov 23 at 14:06
gimusi
93k94495
answered Nov 23 at 14:06
gimusi
93k94495
answered Nov 23 at 14:06
gimusi
93k94495
answered Nov 23 at 14:06
gimusi
93k94495
93k94495
add a comment |
add a comment |
up vote
1
down vote
The error lies in this step:
$|frac{n^2 - 2n}{n+1} - L| <|n-L|$
You know that $frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?
answered Nov 23 at 14:14
Shubham Johri
2,819413
add a comment |
up vote
1
down vote
The error lies in this step:
$|frac{n^2 - 2n}{n+1} - L| <|n-L|$
You know that $frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?
answered Nov 23 at 14:14
Shubham Johri
2,819413
add a comment |
up vote
1
down vote
up vote
1
down vote
The error lies in this step:
$|frac{n^2 - 2n}{n+1} - L| <|n-L|$
You know that $frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?
answered Nov 23 at 14:14
Shubham Johri
2,819413
The error lies in this step:
$|frac{n^2 - 2n}{n+1} - L| <|n-L|$
You know that $frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?
answered Nov 23 at 14:14
Shubham Johri
2,819413
edited Nov 23 at 14:23
answered Nov 23 at 14:14
Shubham Johri
2,819413
answered Nov 23 at 14:14
Shubham Johri
2,819413
answered Nov 23 at 14:14
Shubham Johri
2,819413
2,819413
add a comment |
add a comment |
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You can also use long division to get $frac{n^2-2n}{n+1} = n-3+frac{3}{n+1}$, and then you only need to consider $frac{3}{n+1}$.
– Toby Mak
Nov 23 at 14:16