Mixing
If $(a_n)toinfty$, then $(1/a_n)to0$
2
$begingroup$
Prove: If $(a_n)toinfty$, then $(1/a_n)to0$
As $a_ntoinfty$, for every $C>0$ there exists an $N$ in the natural numbers such that we have
$a_n>C$ whenever $n>N$
$implies|a_n|>C$ whenever $n>N$
$implies|1/a_n| < 1/C$ whenever $n>N$
If follows that if we choose $e>0$ there exists an $N_1$ such that we have
$|1/a_n| < e$ whenever $n>N_1$
How is this proof, is it acceptable do you think?
proof-verification
edited Jan 10 at 17:49
Shubham Johri
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
$endgroup$
add a comment |
2
$begingroup$
Prove: If $(a_n)toinfty$, then $(1/a_n)to0$
As $a_ntoinfty$, for every $C>0$ there exists an $N$ in the natural numbers such that we have
$a_n>C$ whenever $n>N$
$implies|a_n|>C$ whenever $n>N$
$implies|1/a_n| < 1/C$ whenever $n>N$
If follows that if we choose $e>0$ there exists an $N_1$ such that we have
$|1/a_n| < e$ whenever $n>N_1$
How is this proof, is it acceptable do you think?
proof-verification
edited Jan 10 at 17:49
Shubham Johri
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
$endgroup$
1
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56
add a comment |
2
2
2
$begingroup$
Prove: If $(a_n)toinfty$, then $(1/a_n)to0$
As $a_ntoinfty$, for every $C>0$ there exists an $N$ in the natural numbers such that we have
$a_n>C$ whenever $n>N$
$implies|a_n|>C$ whenever $n>N$
$implies|1/a_n| < 1/C$ whenever $n>N$
If follows that if we choose $e>0$ there exists an $N_1$ such that we have
$|1/a_n| < e$ whenever $n>N_1$
How is this proof, is it acceptable do you think?
proof-verification
edited Jan 10 at 17:49
Shubham Johri
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
$endgroup$
Prove: If $(a_n)toinfty$, then $(1/a_n)to0$
As $a_ntoinfty$, for every $C>0$ there exists an $N$ in the natural numbers such that we have
$a_n>C$ whenever $n>N$
$implies|a_n|>C$ whenever $n>N$
$implies|1/a_n| < 1/C$ whenever $n>N$
If follows that if we choose $e>0$ there exists an $N_1$ such that we have
$|1/a_n| < e$ whenever $n>N_1$
How is this proof, is it acceptable do you think?
proof-verification
proof-verification
edited Jan 10 at 17:49
Shubham Johri
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
edited Jan 10 at 17:49
Shubham Johri
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
edited Jan 10 at 17:49
Shubham Johri
5,102717
edited Jan 10 at 17:49
Shubham Johri
5,102717
edited Jan 10 at 17:49
Shubham Johri
5,102717
5,102717
asked Jan 10 at 17:28
James DohertyJames Doherty
527
asked Jan 10 at 17:28
James DohertyJames Doherty
527
asked Jan 10 at 17:28
James DohertyJames Doherty
527
527
1
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56
add a comment |
1
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56
1
1
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56
add a comment |
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1
$begingroup$
Almost. For $varepsilon$ choose $C:=1/varepsilon$
$endgroup$
– Berci
Jan 10 at 17:43
$begingroup$
Thanks I don't quite understand are you saying that for the last bit I should say that if we choose e=1/C the definition will be satisfied?
$endgroup$
– James Doherty
Jan 10 at 17:50
$begingroup$
Yes, I miss that part from the proof.. But not exactly.. First, $varepsilon$ is given, and we're looking for an $N_1$, which will be chosen for $C=1/varepsilon$ by the hypothesis.
$endgroup$
– Berci
Jan 10 at 17:51
$begingroup$
@JamesDoherty You need to show that $forallepsilon>0,exists n_0inBbb N:|1/a_n|<epsilonforall nge n_0$. To prove this, take $C=1/epsilon>0$. Then $exists n_1inBbb N:a_n>Cforall nge n_1implies 1/a_n=|1/a_n|<1/C=epsilonforall nge n_0=n_1$
$endgroup$
– Shubham Johri
Jan 10 at 17:56