Proving that a sequence of reciprocated integers is not Cauchy [on hold]Proving a Sequence is CauchyProving...
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Proving that a sequence of reciprocated integers is not Cauchy [on hold]
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Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.
I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.
real-analysis cauchy-sequences
asked 20 hours ago
B RetnikB Retnik
534
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put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.
I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.
real-analysis cauchy-sequences
asked 20 hours ago
B RetnikB Retnik
534
$endgroup$
put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.
I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.
real-analysis cauchy-sequences
asked 20 hours ago
B RetnikB Retnik
534
$endgroup$
Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.
I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.
real-analysis cauchy-sequences
real-analysis cauchy-sequences
asked 20 hours ago
B RetnikB Retnik
534
asked 20 hours ago
B RetnikB Retnik
534
asked 20 hours ago
B RetnikB Retnik
534
asked 20 hours ago
B RetnikB Retnik
534
asked 20 hours ago
B RetnikB Retnik
534
534
put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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2 Answers
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We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$
Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.
Can you take it from here ?
answered 19 hours ago
FredFred
47.7k1849
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$frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
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$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
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– Kavi Rama Murthy
19 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$
Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.
Can you take it from here ?
answered 19 hours ago
FredFred
47.7k1849
$endgroup$
add a comment |
$begingroup$
We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$
Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.
Can you take it from here ?
answered 19 hours ago
FredFred
47.7k1849
$endgroup$
add a comment |
$begingroup$
We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$
Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.
Can you take it from here ?
answered 19 hours ago
FredFred
47.7k1849
$endgroup$
We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$
Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.
Can you take it from here ?
answered 19 hours ago
FredFred
47.7k1849
answered 19 hours ago
FredFred
47.7k1849
answered 19 hours ago
FredFred
47.7k1849
answered 19 hours ago
FredFred
47.7k1849
47.7k1849
add a comment |
add a comment |
$begingroup$
$frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
add a comment |
$begingroup$
$frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
add a comment |
$begingroup$
$frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 20 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
65.4k42766
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
add a comment |
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
$begingroup$
Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
$endgroup$
– B Retnik
19 hours ago
2
2
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
$begingroup$
Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
$endgroup$
– Kavi Rama Murthy
19 hours ago
add a comment |
