Does $b_n = a_n+1 - a_n $ is a zero sequence implies $a_n$ is convergent? [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite













This question already has an answer here:



  • Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

    2 answers



Let $(a_n)$ be a sequence in $mathbbR$ and let $b_n = a_n+1 - a_n $ be a zero sequence. According to my intuition, I would say that $a_n$ converges. But my solution says otherwise. How could that be? I just can't find a counterexample.




real-analysis sequences-and-series convergence




share|cite|improve this question













marked as duplicate by Martin R, Daniel Fischer♦ real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
 Jul 25 at 11:47


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
    – Lord Shark the Unknown
    Jul 25 at 10:51











  • Another one: math.stackexchange.com/q/1019832/42969.
    – Martin R
    Jul 25 at 11:00










  • You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
    – kingW3
    Jul 25 at 11:09











  • Counterexample $a_n=frac1n$
    – Dr. Wolfgang Hintze
    Jul 25 at 13:10














up vote
0
down vote

favorite













This question already has an answer here:



  • Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

    2 answers



Let $(a_n)$ be a sequence in $mathbbR$ and let $b_n = a_n+1 - a_n $ be a zero sequence. According to my intuition, I would say that $a_n$ converges. But my solution says otherwise. How could that be? I just can't find a counterexample.




real-analysis sequences-and-series convergence




share|cite|improve this question













marked as duplicate by Martin R, Daniel Fischer♦ real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
 Jul 25 at 11:47


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
    – Lord Shark the Unknown
    Jul 25 at 10:51











  • Another one: math.stackexchange.com/q/1019832/42969.
    – Martin R
    Jul 25 at 11:00










  • You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
    – kingW3
    Jul 25 at 11:09











  • Counterexample $a_n=frac1n$
    – Dr. Wolfgang Hintze
    Jul 25 at 13:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite












This question already has an answer here:



  • Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

    2 answers



Let $(a_n)$ be a sequence in $mathbbR$ and let $b_n = a_n+1 - a_n $ be a zero sequence. According to my intuition, I would say that $a_n$ converges. But my solution says otherwise. How could that be? I just can't find a counterexample.




real-analysis sequences-and-series convergence




share|cite|improve this question














This question already has an answer here:



  • Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

    2 answers



Let $(a_n)$ be a sequence in $mathbbR$ and let $b_n = a_n+1 - a_n $ be a zero sequence. According to my intuition, I would say that $a_n$ converges. But my solution says otherwise. How could that be? I just can't find a counterexample.





This question already has an answer here:



  • Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

    2 answers





real-analysis sequences-and-series convergence





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 25 at 11:38









Shaun

7,32592972




7,32592972









asked Jul 25 at 10:50









Anna Saabel

1569




1569




marked as duplicate by Martin R, Daniel Fischer♦ real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
 Jul 25 at 11:47


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Martin R, Daniel Fischer♦ real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
 Jul 25 at 11:47


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
    – Lord Shark the Unknown
    Jul 25 at 10:51











  • Another one: math.stackexchange.com/q/1019832/42969.
    – Martin R
    Jul 25 at 11:00










  • You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
    – kingW3
    Jul 25 at 11:09











  • Counterexample $a_n=frac1n$
    – Dr. Wolfgang Hintze
    Jul 25 at 13:10












  • 1




    A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
    – Lord Shark the Unknown
    Jul 25 at 10:51











  • Another one: math.stackexchange.com/q/1019832/42969.
    – Martin R
    Jul 25 at 11:00










  • You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
    – kingW3
    Jul 25 at 11:09











  • Counterexample $a_n=frac1n$
    – Dr. Wolfgang Hintze
    Jul 25 at 13:10







1




1




A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
– Lord Shark the Unknown
Jul 25 at 10:51





A zero sequence? Is that one that converges to zero? Like say $sqrtn+1-sqrt n$?
– Lord Shark the Unknown
Jul 25 at 10:51













Another one: math.stackexchange.com/q/1019832/42969.
– Martin R
Jul 25 at 11:00




Another one: math.stackexchange.com/q/1019832/42969.
– Martin R
Jul 25 at 11:00












You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
– kingW3
Jul 25 at 11:09





You have that a sequence converges if it's cauchy, since this one is a special case of cauchy $m=n+1$, one would expect that there exists such sequence so that there exists a $m$ such that $a_n-a_m$ doesn't go to $0$. Intuitively, a sequence which diverges slowly will satisfy $a_n+1-a_nto 0$.
– kingW3
Jul 25 at 11:09













Counterexample $a_n=frac1n$
– Dr. Wolfgang Hintze
Jul 25 at 13:10




Counterexample $a_n=frac1n$
– Dr. Wolfgang Hintze
Jul 25 at 13:10










3 Answers
3






active

oldest

votes

















up vote
2
down vote













$$a_n = sum_k=1^n frac 1 k$$






share|cite|improve this answer




























    up vote
    0
    down vote













    Counterexample: $a_n=1+frac12+...+frac1n$.






    share|cite|improve this answer





















    • Why the downvote ???????????????
      – Fred
      Jul 26 at 5:57

















    up vote
    -1
    down vote













    Another nice example. $a_n = sqrtn$. Show that
    $$
    sqrtn+1 - sqrtn to 0qquadtextbutqquad
    sqrtn to +infty
    $$






    share|cite|improve this answer




























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote













      $$a_n = sum_k=1^n frac 1 k$$






      share|cite|improve this answer

























        up vote
        2
        down vote













        $$a_n = sum_k=1^n frac 1 k$$






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          $$a_n = sum_k=1^n frac 1 k$$






          share|cite|improve this answer













          $$a_n = sum_k=1^n frac 1 k$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 25 at 10:52









          Kenny Lau

          18.5k2157




          18.5k2157




















              up vote
              0
              down vote













              Counterexample: $a_n=1+frac12+...+frac1n$.






              share|cite|improve this answer





















              • Why the downvote ???????????????
                – Fred
                Jul 26 at 5:57














              up vote
              0
              down vote













              Counterexample: $a_n=1+frac12+...+frac1n$.






              share|cite|improve this answer





















              • Why the downvote ???????????????
                – Fred
                Jul 26 at 5:57












              up vote
              0
              down vote










              up vote
              0
              down vote









              Counterexample: $a_n=1+frac12+...+frac1n$.






              share|cite|improve this answer













              Counterexample: $a_n=1+frac12+...+frac1n$.







              share|cite|improve this answer













              share|cite|improve this answer



              share|cite|improve this answer











              answered Jul 25 at 10:52









              Fred

              37.2k1237




              37.2k1237











              • Why the downvote ???????????????
                – Fred
                Jul 26 at 5:57
















              • Why the downvote ???????????????
                – Fred
                Jul 26 at 5:57















              Why the downvote ???????????????
              – Fred
              Jul 26 at 5:57




              Why the downvote ???????????????
              – Fred
              Jul 26 at 5:57










              up vote
              -1
              down vote













              Another nice example. $a_n = sqrtn$. Show that
              $$
              sqrtn+1 - sqrtn to 0qquadtextbutqquad
              sqrtn to +infty
              $$






              share|cite|improve this answer

























                up vote
                -1
                down vote













                Another nice example. $a_n = sqrtn$. Show that
                $$
                sqrtn+1 - sqrtn to 0qquadtextbutqquad
                sqrtn to +infty
                $$






                share|cite|improve this answer























                  up vote
                  -1
                  down vote










                  up vote
                  -1
                  down vote









                  Another nice example. $a_n = sqrtn$. Show that
                  $$
                  sqrtn+1 - sqrtn to 0qquadtextbutqquad
                  sqrtn to +infty
                  $$






                  share|cite|improve this answer













                  Another nice example. $a_n = sqrtn$. Show that
                  $$
                  sqrtn+1 - sqrtn to 0qquadtextbutqquad
                  sqrtn to +infty
                  $$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 25 at 11:23









                  GEdgar

                  58.4k264163




                  58.4k264163












                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more