If $ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0 $ then is it a Cauchy sequence? [on hold] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?Series constructed from a cauchy sequenceShow that for a sequence of real numbers $(a_n)_n$ $lim_n a_n=0$ implies $frac1nsum_i=0^n-1lvert a_irvert=0$Show a sequence such that $lim_ N to infty sum_n=1^N lvert a_n-a_n+1rvert< infty$, is CauchyWhy does $lim_k rightarrow infty lvert k^2 sin (k^4) rvert = 0$?Can $(a_n)_n$ with $limsuplimits_nrightarrowinftyleftlvert fraca_n+1a_n rightrvert = infty$ be a null sequence?Prove: The limit of a Cauchy sequence $a_n$ = $lim_ntoinftya_n$$ lim_nrightarrow infty[a_n-a]=0 $ implies $ lim_nrightarrow inftyfrac1n[a_n-a]=0 $?$lim_x rightarrow a lvert f(x)rvert = lvertlim_x rightarrow a f(x)rvert$?Prove $ lim_ntoinftylVert x^nrVert^frac1n = inf_ngeq 1lVert x^nrVert^frac1 n$ by Fekete’s LemmaFinding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$

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If $ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0 $ then is it a Cauchy sequence? [on hold]



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?Series constructed from a cauchy sequenceShow that for a sequence of real numbers $(a_n)_n$ $lim_n a_n=0$ implies $frac1nsum_i=0^n-1lvert a_irvert=0$Show a sequence such that $lim_ N to infty sum_n=1^N lvert a_n-a_n+1rvert< infty$, is CauchyWhy does $lim_k rightarrow infty lvert k^2 sin (k^4) rvert = 0$?Can $(a_n)_n$ with $limsuplimits_nrightarrowinftyleftlvert fraca_n+1a_n rightrvert = infty$ be a null sequence?Prove: The limit of a Cauchy sequence $a_n$ = $lim_ntoinftya_n$$ lim_nrightarrow infty[a_n-a]=0 $ implies $ lim_nrightarrow inftyfrac1n[a_n-a]=0 $?$lim_x rightarrow a lvert f(x)rvert = lvertlim_x rightarrow a f(x)rvert$?Prove $ lim_ntoinftylVert x^nrVert^frac1n = inf_ngeq 1lVert x^nrVert^frac1 n$ by Fekete’s LemmaFinding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$










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Let $(a_n)$ be a sequence of real numbers, for which it holds, that
$$ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?










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$endgroup$



put on hold as off-topic by user21820, Saad, RRL, max_zorn, Lee David Chung Lin 2 days 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." – user21820, Saad, RRL, max_zorn, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 13:36















1












$begingroup$


Let $(a_n)$ be a sequence of real numbers, for which it holds, that
$$ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?










share|cite|improve this question











$endgroup$



put on hold as off-topic by user21820, Saad, RRL, max_zorn, Lee David Chung Lin 2 days 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." – user21820, Saad, RRL, max_zorn, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 13:36













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$begingroup$


Let $(a_n)$ be a sequence of real numbers, for which it holds, that
$$ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?










share|cite|improve this question











$endgroup$




Let $(a_n)$ be a sequence of real numbers, for which it holds, that
$$ lim_n rightarrow infty lvert a_n+1-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?







limits convergence cauchy-sequences






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edited Apr 15 at 6:28









user21820

40.4k544163




40.4k544163










asked Apr 14 at 23:25









Joker123Joker123

756313




756313




put on hold as off-topic by user21820, Saad, RRL, max_zorn, Lee David Chung Lin 2 days 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." – user21820, Saad, RRL, max_zorn, Lee David Chung Lin
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 user21820, Saad, RRL, max_zorn, Lee David Chung Lin 2 days 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." – user21820, Saad, RRL, max_zorn, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 13:36
















  • $begingroup$
    Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
    $endgroup$
    – YuiTo Cheng
    Apr 15 at 13:36















$begingroup$
Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
$endgroup$
– YuiTo Cheng
Apr 15 at 13:36




$begingroup$
Possible duplicate of Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
$endgroup$
– YuiTo Cheng
Apr 15 at 13:36










3 Answers
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Unfortunately not. Consider
$$a_n:=sum_i=1^nfrac1i.$$
We find $a_n+1-a_n=1/(n+1)to 0,$ but $lim_ntoinftya_n=infty,$ hence $a_n_ninmathbbN$ is not a cauchy sequence.






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    Counterexample: $a_n = sqrtn$. Clearly this sequence does not converge. But
    $$
    a_n+1 - a_n = sqrtn+1 - sqrtn = frac(sqrtn+1 - sqrtn)(sqrtn+1 + sqrtn)(sqrtn+1 + sqrtn) = frac1sqrtn+1 + sqrtn to 0 , .
    $$






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      No. The sequence $a_n=sum_k=1^nfrac1k$ is a counterexample.






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        3 Answers
        3






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        3 Answers
        3






        active

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        active

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        active

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        $begingroup$

        Unfortunately not. Consider
        $$a_n:=sum_i=1^nfrac1i.$$
        We find $a_n+1-a_n=1/(n+1)to 0,$ but $lim_ntoinftya_n=infty,$ hence $a_n_ninmathbbN$ is not a cauchy sequence.






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        $endgroup$

















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          $begingroup$

          Unfortunately not. Consider
          $$a_n:=sum_i=1^nfrac1i.$$
          We find $a_n+1-a_n=1/(n+1)to 0,$ but $lim_ntoinftya_n=infty,$ hence $a_n_ninmathbbN$ is not a cauchy sequence.






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          $endgroup$















            2












            2








            2





            $begingroup$

            Unfortunately not. Consider
            $$a_n:=sum_i=1^nfrac1i.$$
            We find $a_n+1-a_n=1/(n+1)to 0,$ but $lim_ntoinftya_n=infty,$ hence $a_n_ninmathbbN$ is not a cauchy sequence.






            share|cite|improve this answer











            $endgroup$



            Unfortunately not. Consider
            $$a_n:=sum_i=1^nfrac1i.$$
            We find $a_n+1-a_n=1/(n+1)to 0,$ but $lim_ntoinftya_n=infty,$ hence $a_n_ninmathbbN$ is not a cauchy sequence.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 14 at 23:33









            HAMIDINE SOUMARE

            2,775418




            2,775418










            answered Apr 14 at 23:28









            MelodyMelody

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                $begingroup$

                Counterexample: $a_n = sqrtn$. Clearly this sequence does not converge. But
                $$
                a_n+1 - a_n = sqrtn+1 - sqrtn = frac(sqrtn+1 - sqrtn)(sqrtn+1 + sqrtn)(sqrtn+1 + sqrtn) = frac1sqrtn+1 + sqrtn to 0 , .
                $$






                share|cite|improve this answer









                $endgroup$

















                  4












                  $begingroup$

                  Counterexample: $a_n = sqrtn$. Clearly this sequence does not converge. But
                  $$
                  a_n+1 - a_n = sqrtn+1 - sqrtn = frac(sqrtn+1 - sqrtn)(sqrtn+1 + sqrtn)(sqrtn+1 + sqrtn) = frac1sqrtn+1 + sqrtn to 0 , .
                  $$






                  share|cite|improve this answer









                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    Counterexample: $a_n = sqrtn$. Clearly this sequence does not converge. But
                    $$
                    a_n+1 - a_n = sqrtn+1 - sqrtn = frac(sqrtn+1 - sqrtn)(sqrtn+1 + sqrtn)(sqrtn+1 + sqrtn) = frac1sqrtn+1 + sqrtn to 0 , .
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    Counterexample: $a_n = sqrtn$. Clearly this sequence does not converge. But
                    $$
                    a_n+1 - a_n = sqrtn+1 - sqrtn = frac(sqrtn+1 - sqrtn)(sqrtn+1 + sqrtn)(sqrtn+1 + sqrtn) = frac1sqrtn+1 + sqrtn to 0 , .
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 14 at 23:49









                    Hans EnglerHans Engler

                    10.8k11936




                    10.8k11936





















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                        No. The sequence $a_n=sum_k=1^nfrac1k$ is a counterexample.






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                          2












                          $begingroup$

                          No. The sequence $a_n=sum_k=1^nfrac1k$ is a counterexample.






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                            2












                            2








                            2





                            $begingroup$

                            No. The sequence $a_n=sum_k=1^nfrac1k$ is a counterexample.






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                            $endgroup$



                            No. The sequence $a_n=sum_k=1^nfrac1k$ is a counterexample.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 14 at 23:28









                            MarkMark

                            11.1k1824




                            11.1k1824













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