Proof that a sequence is Cauchy.












4














Show that $left( x_{n}right) $ is a Cauchy sequence, where
$$
x_{n}=frac{sin1}{2}+frac{sin2}{2^{2}}+ldots+frac{sin n}{2^{n}}.
$$



We try to evaluate $leftvert x_{n+p}-x_{n}rightvert $ and to show that
this one is arbitrary small. So
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{sinleft(
n+1right) }{2^{n+1}}+frac{sinleft( n+2right) }{2^{n+2}}+ldots
+frac{sinleft( n+pright) }{2^{n+p}}rightvert \
& leqfrac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}.
end{align*}
Now, $left( frac{1}{2^{n+j}}right) $ converge all to zero. Hence
$leftvert a_{n+p}-a_{n}rightvert rightarrow0$.
Does it work this argument according to the definition of Cauchy sequence
$$
forallvarepsilon>0,
quad
exists n_{varepsilon}in
mathbb{N}
,
quad
forall nin
mathbb{N}
,
quad
forall pin
mathbb{N}
:n,pgeq n_{varepsilon}Rightarrowleftvert a_{n+p}-a_{n}rightvert
<varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+frac{1}{2}+ldots+frac{1}{n},
$$
we have that
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}rightvert \
& leqfrac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}
end{align*}
and also all sequences $left( frac{1}{n+j}right) _{n}$ converge to zero,
but this time $left( a_{n}right) $ is not a Cauchy sequence.






analysis







share|cite|improve this question




















  • 1




    the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
    – Offlaw
    Jan 4 at 14:16






  • 1




    an infinite sum of terms converging to zero may not converge to zero
    – Smilia
    Jan 4 at 14:16
















4














Show that $left( x_{n}right) $ is a Cauchy sequence, where
$$
x_{n}=frac{sin1}{2}+frac{sin2}{2^{2}}+ldots+frac{sin n}{2^{n}}.
$$



We try to evaluate $leftvert x_{n+p}-x_{n}rightvert $ and to show that
this one is arbitrary small. So
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{sinleft(
n+1right) }{2^{n+1}}+frac{sinleft( n+2right) }{2^{n+2}}+ldots
+frac{sinleft( n+pright) }{2^{n+p}}rightvert \
& leqfrac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}.
end{align*}
Now, $left( frac{1}{2^{n+j}}right) $ converge all to zero. Hence
$leftvert a_{n+p}-a_{n}rightvert rightarrow0$.
Does it work this argument according to the definition of Cauchy sequence
$$
forallvarepsilon>0,
quad
exists n_{varepsilon}in
mathbb{N}
,
quad
forall nin
mathbb{N}
,
quad
forall pin
mathbb{N}
:n,pgeq n_{varepsilon}Rightarrowleftvert a_{n+p}-a_{n}rightvert
<varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+frac{1}{2}+ldots+frac{1}{n},
$$
we have that
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}rightvert \
& leqfrac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}
end{align*}
and also all sequences $left( frac{1}{n+j}right) _{n}$ converge to zero,
but this time $left( a_{n}right) $ is not a Cauchy sequence.






analysis







share|cite|improve this question




















  • 1




    the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
    – Offlaw
    Jan 4 at 14:16






  • 1




    an infinite sum of terms converging to zero may not converge to zero
    – Smilia
    Jan 4 at 14:16














4












4








4


1





Show that $left( x_{n}right) $ is a Cauchy sequence, where
$$
x_{n}=frac{sin1}{2}+frac{sin2}{2^{2}}+ldots+frac{sin n}{2^{n}}.
$$



We try to evaluate $leftvert x_{n+p}-x_{n}rightvert $ and to show that
this one is arbitrary small. So
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{sinleft(
n+1right) }{2^{n+1}}+frac{sinleft( n+2right) }{2^{n+2}}+ldots
+frac{sinleft( n+pright) }{2^{n+p}}rightvert \
& leqfrac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}.
end{align*}
Now, $left( frac{1}{2^{n+j}}right) $ converge all to zero. Hence
$leftvert a_{n+p}-a_{n}rightvert rightarrow0$.
Does it work this argument according to the definition of Cauchy sequence
$$
forallvarepsilon>0,
quad
exists n_{varepsilon}in
mathbb{N}
,
quad
forall nin
mathbb{N}
,
quad
forall pin
mathbb{N}
:n,pgeq n_{varepsilon}Rightarrowleftvert a_{n+p}-a_{n}rightvert
<varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+frac{1}{2}+ldots+frac{1}{n},
$$
we have that
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}rightvert \
& leqfrac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}
end{align*}
and also all sequences $left( frac{1}{n+j}right) _{n}$ converge to zero,
but this time $left( a_{n}right) $ is not a Cauchy sequence.






analysis







share|cite|improve this question















Show that $left( x_{n}right) $ is a Cauchy sequence, where
$$
x_{n}=frac{sin1}{2}+frac{sin2}{2^{2}}+ldots+frac{sin n}{2^{n}}.
$$



We try to evaluate $leftvert x_{n+p}-x_{n}rightvert $ and to show that
this one is arbitrary small. So
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{sinleft(
n+1right) }{2^{n+1}}+frac{sinleft( n+2right) }{2^{n+2}}+ldots
+frac{sinleft( n+pright) }{2^{n+p}}rightvert \
& leqfrac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}.
end{align*}
Now, $left( frac{1}{2^{n+j}}right) $ converge all to zero. Hence
$leftvert a_{n+p}-a_{n}rightvert rightarrow0$.
Does it work this argument according to the definition of Cauchy sequence
$$
forallvarepsilon>0,
quad
exists n_{varepsilon}in
mathbb{N}
,
quad
forall nin
mathbb{N}
,
quad
forall pin
mathbb{N}
:n,pgeq n_{varepsilon}Rightarrowleftvert a_{n+p}-a_{n}rightvert
<varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+frac{1}{2}+ldots+frac{1}{n},
$$
we have that
begin{align*}
leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}rightvert \
& leqfrac{1}{n+1}+frac{1}{n+2}+ldots+frac{1}{n+p}
end{align*}
and also all sequences $left( frac{1}{n+j}right) _{n}$ converge to zero,
but this time $left( a_{n}right) $ is not a Cauchy sequence.






analysis




analysis






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share|cite|improve this question













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edited Jan 4 at 14:08







stefano

















asked Jan 4 at 13:59









stefanostefano

824




824








  • 1




    the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
    – Offlaw
    Jan 4 at 14:16






  • 1




    an infinite sum of terms converging to zero may not converge to zero
    – Smilia
    Jan 4 at 14:16














  • 1




    the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
    – Offlaw
    Jan 4 at 14:16






  • 1




    an infinite sum of terms converging to zero may not converge to zero
    – Smilia
    Jan 4 at 14:16








1




1




the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
– Offlaw
Jan 4 at 14:16




the sequence converges, but not because of $frac{1}{2^{n}} to 0$ as $n to infty$. Use sum of G.P.
– Offlaw
Jan 4 at 14:16




1




1




an infinite sum of terms converging to zero may not converge to zero
– Smilia
Jan 4 at 14:16




an infinite sum of terms converging to zero may not converge to zero
– Smilia
Jan 4 at 14:16










2 Answers
2






active

oldest

votes


















2














It is not sufficient that all $left( frac{1}{2^{n+j}}right)$ converge to zero. A correct argument would be that
$$
leftvert a_{n+p}-a_{n}rightvert le
frac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}
= frac{1}{2^{n}} left( frac 12 + frac 14 + ldots + frac{1}{2^{p}}right) < frac{1}{2^{n}}
$$
becomes arbitrary small for large $n$ and arbitrary $p$.



That won't work for the harmonic series.






share|cite|improve this answer





























    0














    A (quite silly) argument...



    $x_n$ is the partial sum of an absolutely converging real sequence. Hence $(x_n)$ converges and is Cauchy as $mathbb R$ is complete.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      It is not sufficient that all $left( frac{1}{2^{n+j}}right)$ converge to zero. A correct argument would be that
      $$
      leftvert a_{n+p}-a_{n}rightvert le
      frac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}
      = frac{1}{2^{n}} left( frac 12 + frac 14 + ldots + frac{1}{2^{p}}right) < frac{1}{2^{n}}
      $$
      becomes arbitrary small for large $n$ and arbitrary $p$.



      That won't work for the harmonic series.






      share|cite|improve this answer


























        2














        It is not sufficient that all $left( frac{1}{2^{n+j}}right)$ converge to zero. A correct argument would be that
        $$
        leftvert a_{n+p}-a_{n}rightvert le
        frac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}
        = frac{1}{2^{n}} left( frac 12 + frac 14 + ldots + frac{1}{2^{p}}right) < frac{1}{2^{n}}
        $$
        becomes arbitrary small for large $n$ and arbitrary $p$.



        That won't work for the harmonic series.






        share|cite|improve this answer
























          2












          2








          2






          It is not sufficient that all $left( frac{1}{2^{n+j}}right)$ converge to zero. A correct argument would be that
          $$
          leftvert a_{n+p}-a_{n}rightvert le
          frac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}
          = frac{1}{2^{n}} left( frac 12 + frac 14 + ldots + frac{1}{2^{p}}right) < frac{1}{2^{n}}
          $$
          becomes arbitrary small for large $n$ and arbitrary $p$.



          That won't work for the harmonic series.






          share|cite|improve this answer












          It is not sufficient that all $left( frac{1}{2^{n+j}}right)$ converge to zero. A correct argument would be that
          $$
          leftvert a_{n+p}-a_{n}rightvert le
          frac{1}{2^{n+1}}+frac{1}{2^{n+2}}+ldots+frac{1}{2^{n+p}}
          = frac{1}{2^{n}} left( frac 12 + frac 14 + ldots + frac{1}{2^{p}}right) < frac{1}{2^{n}}
          $$
          becomes arbitrary small for large $n$ and arbitrary $p$.



          That won't work for the harmonic series.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 14:19









          Martin RMartin R

          27.3k33254




          27.3k33254























              0














              A (quite silly) argument...



              $x_n$ is the partial sum of an absolutely converging real sequence. Hence $(x_n)$ converges and is Cauchy as $mathbb R$ is complete.






              share|cite|improve this answer


























                0














                A (quite silly) argument...



                $x_n$ is the partial sum of an absolutely converging real sequence. Hence $(x_n)$ converges and is Cauchy as $mathbb R$ is complete.






                share|cite|improve this answer
























                  0












                  0








                  0






                  A (quite silly) argument...



                  $x_n$ is the partial sum of an absolutely converging real sequence. Hence $(x_n)$ converges and is Cauchy as $mathbb R$ is complete.






                  share|cite|improve this answer












                  A (quite silly) argument...



                  $x_n$ is the partial sum of an absolutely converging real sequence. Hence $(x_n)$ converges and is Cauchy as $mathbb R$ is complete.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 14:14









                  mathcounterexamples.netmathcounterexamples.net

                  25.3k21953




                  25.3k21953






























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