Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?












3












$begingroup$


If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$



and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,



Then can we conclude that $a_n$ is convergent?



$$space$$
I know that without condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467



$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$



This oscillates between 0 and 1, while the difference of consecutive terms approaches 0 since the difference is always of the form $pmfrac1m$ and m increases the further we go in this sequence.



So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?






real-analysis sequences-and-series







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  • 1




    $begingroup$
    Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
    $endgroup$
    – Adayah
    38 mins ago
















3












$begingroup$


If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$



and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,



Then can we conclude that $a_n$ is convergent?



$$space$$
I know that without condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467



$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$



This oscillates between 0 and 1, while the difference of consecutive terms approaches 0 since the difference is always of the form $pmfrac1m$ and m increases the further we go in this sequence.



So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?






real-analysis sequences-and-series







share|cite|improve this question







New contributor




M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
    $endgroup$
    – Adayah
    38 mins ago














3












3








3


1



$begingroup$


If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$



and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,



Then can we conclude that $a_n$ is convergent?



$$space$$
I know that without condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467



$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$



This oscillates between 0 and 1, while the difference of consecutive terms approaches 0 since the difference is always of the form $pmfrac1m$ and m increases the further we go in this sequence.



So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?






real-analysis sequences-and-series







share|cite|improve this question







New contributor




M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




If $a_n$ is a sequence such that $$a_1 leq a_2 leq a_3 leq ...$$



and has the property that $space$$a_{n+1}-a_n longrightarrow 0$,



Then can we conclude that $a_n$ is convergent?



$$space$$
I know that without condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing: $space$ https://math.stackexchange.com/a/1437395/625467



$0, 1, frac12, 0, frac13, frac23, 1, frac34, frac12, frac14, 0, frac15, frac25, frac35, frac45, 1...$



This oscillates between 0 and 1, while the difference of consecutive terms approaches 0 since the difference is always of the form $pmfrac1m$ and m increases the further we go in this sequence.



So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge? Or is this still not sufficient?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question







New contributor




M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




M D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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asked 2 hours ago









M DM D

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213




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New contributor





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Check out our Code of Conduct.








  • 1




    $begingroup$
    Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
    $endgroup$
    – Adayah
    38 mins ago














  • 1




    $begingroup$
    Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
    $endgroup$
    – Adayah
    38 mins ago








1




1




$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
$endgroup$
– Adayah
38 mins ago




$begingroup$
Note that while your sequence goes up and down periodically, you could define another sequence with the same step length for each $n$ but with all steps positive. That would be a counterexample to your question.
$endgroup$
– Adayah
38 mins ago










3 Answers
3






active

oldest

votes


















14












$begingroup$

No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
    $endgroup$
    – Rhys Hughes
    2 hours ago






  • 3




    $begingroup$
    Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
    $endgroup$
    – M D
    2 hours ago






  • 4




    $begingroup$
    @RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
    $endgroup$
    – Robert Z
    2 hours ago





















5












$begingroup$

An increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ even if we know that $a_{n+1}-a_nto 0$.



An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.



On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.



So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Robert.A fine answer!+
    $endgroup$
    – Peter Szilas
    2 hours ago



















0












$begingroup$

The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.



Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
    $endgroup$
    – Nate Eldredge
    1 hour ago












  • $begingroup$
    @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
    $endgroup$
    – Haris Gusic
    1 hour ago










  • $begingroup$
    Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
    $endgroup$
    – Nate Eldredge
    1 hour ago










  • $begingroup$
    @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
    $endgroup$
    – Haris Gusic
    1 hour ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









14












$begingroup$

No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
    $endgroup$
    – Rhys Hughes
    2 hours ago






  • 3




    $begingroup$
    Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
    $endgroup$
    – M D
    2 hours ago






  • 4




    $begingroup$
    @RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
    $endgroup$
    – Robert Z
    2 hours ago


















14












$begingroup$

No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
    $endgroup$
    – Rhys Hughes
    2 hours ago






  • 3




    $begingroup$
    Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
    $endgroup$
    – M D
    2 hours ago






  • 4




    $begingroup$
    @RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
    $endgroup$
    – Robert Z
    2 hours ago
















14












14








14





$begingroup$

No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$.






share|cite|improve this answer











$endgroup$



No. Just consider the case in which $a_n=1+frac12+frac13+cdots+frac1n$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 16 mins ago









Riker

1055




1055










answered 2 hours ago









José Carlos SantosJosé Carlos Santos

162k22128233




162k22128233












  • $begingroup$
    Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
    $endgroup$
    – Rhys Hughes
    2 hours ago






  • 3




    $begingroup$
    Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
    $endgroup$
    – M D
    2 hours ago






  • 4




    $begingroup$
    @RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
    $endgroup$
    – Robert Z
    2 hours ago




















  • $begingroup$
    Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
    $endgroup$
    – Rhys Hughes
    2 hours ago






  • 3




    $begingroup$
    Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
    $endgroup$
    – M D
    2 hours ago






  • 4




    $begingroup$
    @RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
    $endgroup$
    – Robert Z
    2 hours ago


















$begingroup$
Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
$endgroup$
– Rhys Hughes
2 hours ago




$begingroup$
Ummm, why is this relevant? Not only are we talking about sequences, not series, but also your series has decreasing terms, which the OP states shouldn't be the case...
$endgroup$
– Rhys Hughes
2 hours ago




3




3




$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
2 hours ago




$begingroup$
Rhys: His sequence is $1, frac32, frac{11}6, frac{25}{12},...$. Essentially the sequence of partial sums associated with the harmonic series. This is still a sequence, not a series, since it is a finite sum.
$endgroup$
– M D
2 hours ago




4




4




$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
$endgroup$
– Robert Z
2 hours ago






$begingroup$
@RhysHughes $a_n=1+frac12+cdots+frac1n$ IS increasing and $a_{n+1}-a_n=frac{1}{n+1}to 0$.
$endgroup$
– Robert Z
2 hours ago













5












$begingroup$

An increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ even if we know that $a_{n+1}-a_nto 0$.



An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.



On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.



So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Robert.A fine answer!+
    $endgroup$
    – Peter Szilas
    2 hours ago
















5












$begingroup$

An increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ even if we know that $a_{n+1}-a_nto 0$.



An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.



On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.



So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Robert.A fine answer!+
    $endgroup$
    – Peter Szilas
    2 hours ago














5












5








5





$begingroup$

An increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ even if we know that $a_{n+1}-a_nto 0$.



An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.



On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.



So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.






share|cite|improve this answer











$endgroup$



An increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ even if we know that $a_{n+1}-a_nto 0$.



An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.



On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.



So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Robert ZRobert Z

98.1k1066137




98.1k1066137












  • $begingroup$
    Robert.A fine answer!+
    $endgroup$
    – Peter Szilas
    2 hours ago


















  • $begingroup$
    Robert.A fine answer!+
    $endgroup$
    – Peter Szilas
    2 hours ago
















$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
2 hours ago




$begingroup$
Robert.A fine answer!+
$endgroup$
– Peter Szilas
2 hours ago











0












$begingroup$

The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.



Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
    $endgroup$
    – Nate Eldredge
    1 hour ago












  • $begingroup$
    @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
    $endgroup$
    – Haris Gusic
    1 hour ago










  • $begingroup$
    Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
    $endgroup$
    – Nate Eldredge
    1 hour ago










  • $begingroup$
    @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
    $endgroup$
    – Haris Gusic
    1 hour ago
















0












$begingroup$

The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.



Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
    $endgroup$
    – Nate Eldredge
    1 hour ago












  • $begingroup$
    @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
    $endgroup$
    – Haris Gusic
    1 hour ago










  • $begingroup$
    Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
    $endgroup$
    – Nate Eldredge
    1 hour ago










  • $begingroup$
    @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
    $endgroup$
    – Haris Gusic
    1 hour ago














0












0








0





$begingroup$

The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.



Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.






share|cite|improve this answer











$endgroup$



The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.



Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Haris GusicHaris Gusic

17910




17910












  • $begingroup$
    Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
    $endgroup$
    – Nate Eldredge
    1 hour ago












  • $begingroup$
    @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
    $endgroup$
    – Haris Gusic
    1 hour ago










  • $begingroup$
    Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
    $endgroup$
    – Nate Eldredge
    1 hour ago










  • $begingroup$
    @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
    $endgroup$
    – Haris Gusic
    1 hour ago


















  • $begingroup$
    Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
    $endgroup$
    – Nate Eldredge
    1 hour ago












  • $begingroup$
    @NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
    $endgroup$
    – Haris Gusic
    1 hour ago










  • $begingroup$
    Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
    $endgroup$
    – Nate Eldredge
    1 hour ago










  • $begingroup$
    @NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
    $endgroup$
    – Haris Gusic
    1 hour ago
















$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
$endgroup$
– Nate Eldredge
1 hour ago






$begingroup$
Your stated condition $lim_{n to infty} (a_{n+m}-a_n) = 0$ for each $m$ is not equivalent to the Cauchy property, and it does not imply that the sequence $a_n$ converges. Consider $a_n = log n$. I think what you would want is that $lim_{n to infty} (a_{n+m}-a_n) = 0$ uniformly in $m$.
$endgroup$
– Nate Eldredge
1 hour ago














$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
1 hour ago




$begingroup$
@NateEldredge But if we take $m=n$, then the condition is not satisfied, is it?
$endgroup$
– Haris Gusic
1 hour ago












$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
1 hour ago




$begingroup$
Okay, the revised condition (where $m$ is a function of $n$) is correct, though it seems awkward to work with in practice.
$endgroup$
– Nate Eldredge
1 hour ago












$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
1 hour ago




$begingroup$
@NateEldredge I formulated it that way because I am more familiar with it. Sorry about any confusion I might have caused.
$endgroup$
– Haris Gusic
1 hour ago










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