If ${a_n}$ is bounded above and $a_{n+1} - a_{n} > -frac{1}{n^{2}}$, then is ${a_n}$ convergent?












0












$begingroup$


Suppose that a sequence ${a_n}$ is bounded above and if satisfies the condition $a_{n+1} - a_{n} > frac{-1}{n^{2}},$ where $n$ is a natural number. Then ${a_n}$ is convergent. $($true or false$)?$



$a_{n+1} - a_{n} > frac{-1}{n^{2}}$



$a_n - a_{n+1} < frac{1}{n^{2}}$



So, this sequence is monotonically decreasing.



And $$lim_{n to infty} (a_n - a_{n-1})= 0$$



How to proceed to see, if the sequence converges$?$






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
    $endgroup$
    – TonyK
    Dec 27 '18 at 14:35
















0












$begingroup$


Suppose that a sequence ${a_n}$ is bounded above and if satisfies the condition $a_{n+1} - a_{n} > frac{-1}{n^{2}},$ where $n$ is a natural number. Then ${a_n}$ is convergent. $($true or false$)?$



$a_{n+1} - a_{n} > frac{-1}{n^{2}}$



$a_n - a_{n+1} < frac{1}{n^{2}}$



So, this sequence is monotonically decreasing.



And $$lim_{n to infty} (a_n - a_{n-1})= 0$$



How to proceed to see, if the sequence converges$?$






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
    $endgroup$
    – TonyK
    Dec 27 '18 at 14:35














0












0








0





$begingroup$


Suppose that a sequence ${a_n}$ is bounded above and if satisfies the condition $a_{n+1} - a_{n} > frac{-1}{n^{2}},$ where $n$ is a natural number. Then ${a_n}$ is convergent. $($true or false$)?$



$a_{n+1} - a_{n} > frac{-1}{n^{2}}$



$a_n - a_{n+1} < frac{1}{n^{2}}$



So, this sequence is monotonically decreasing.



And $$lim_{n to infty} (a_n - a_{n-1})= 0$$



How to proceed to see, if the sequence converges$?$






real-analysis







share|cite|improve this question











$endgroup$




Suppose that a sequence ${a_n}$ is bounded above and if satisfies the condition $a_{n+1} - a_{n} > frac{-1}{n^{2}},$ where $n$ is a natural number. Then ${a_n}$ is convergent. $($true or false$)?$



$a_{n+1} - a_{n} > frac{-1}{n^{2}}$



$a_n - a_{n+1} < frac{1}{n^{2}}$



So, this sequence is monotonically decreasing.



And $$lim_{n to infty} (a_n - a_{n-1})= 0$$



How to proceed to see, if the sequence converges$?$






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 13:29









StubbornAtom

6,17811339




6,17811339










asked Dec 27 '18 at 13:24









MathsaddictMathsaddict

3669




3669












  • $begingroup$
    It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
    $endgroup$
    – TonyK
    Dec 27 '18 at 14:35


















  • $begingroup$
    It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
    $endgroup$
    – TonyK
    Dec 27 '18 at 14:35
















$begingroup$
It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
$endgroup$
– TonyK
Dec 27 '18 at 14:35




$begingroup$
It's not necessarily monitonically decreasing. There is nothing in the conditions to imply that $a_{n+1}-a_n$ can't be greater than $0$.
$endgroup$
– TonyK
Dec 27 '18 at 14:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

$$frac{1}{n^2} = b_n-b_{n-1}, ngeq2, b_1 = 1 $$
then
$$b_n = sum_{k=1}^n frac{1}{k^2} $$
Then the following sequence is convergent, because it's increasing and bounded:
$$x_n = b_n+a_n $$
Hence $a_n = x_n-b_n$ is also convergent.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    First of all $a_{n}-a_{n+1}<frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
    For example the sequence $a_{n}=n$ satisfy the inequality.



    In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.



    The proof of your question is that: the series $1/n^{2}$ converges, so $forallvarepsilon>0,exists n_{varepsilon}$ tc$forall N>n_{varepsilon}$ $sum_{n=N}^{infty}1/n^{2}<varepsilon$.



    Because the sequence is upper bounded we can call $M:=limsup_{ntoinfty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $forallvarepsilon>0$ $exists k_{varepsilon}$ tc $|M-a_{n_{k_{varepsilon}}}|<varepsilon$.



    From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<sum_{i=n_{k}}^{n}1/i^{2}<sum_{i=n_{k}}^{infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-sum_{i=n_{k}}^{infty}1/i^{2}$.
    It follows that for $k>max{k_{varepsilon},n_{varepsilon}}$ $a_{n}>M-2varepsilon$ $forall n>n_k$.



    On the other hand let $M_{k}:=sup_{n>n_k}{a_{n}}$, so we have that obviously $a_{n}<M_{k}$ $forall n>k$, and it's clear because of the definition that $M_{k}to M$, and it does it decreasingly.
    So for $k>m_{varepsilon}$ we have $M_{k}<M+varepsilon$.



    So for $k>max{k_{varepsilon},n_{varepsilon}, m_{varepsilon}}$ and for $n>n_{k}$ we have $M+varepsilon>a_{n}>M-2varepsilon$, that means $a_nto M$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
      $endgroup$
      – Jakobian
      Dec 27 '18 at 15:21










    • $begingroup$
      You are right, i've changed my answer, hope it's right now
      $endgroup$
      – ecrin
      Dec 27 '18 at 16:55










    • $begingroup$
      I think it's ok
      $endgroup$
      – Jakobian
      Dec 27 '18 at 17:49











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    2 Answers
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    2 Answers
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    2












    $begingroup$

    $$frac{1}{n^2} = b_n-b_{n-1}, ngeq2, b_1 = 1 $$
    then
    $$b_n = sum_{k=1}^n frac{1}{k^2} $$
    Then the following sequence is convergent, because it's increasing and bounded:
    $$x_n = b_n+a_n $$
    Hence $a_n = x_n-b_n$ is also convergent.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $$frac{1}{n^2} = b_n-b_{n-1}, ngeq2, b_1 = 1 $$
      then
      $$b_n = sum_{k=1}^n frac{1}{k^2} $$
      Then the following sequence is convergent, because it's increasing and bounded:
      $$x_n = b_n+a_n $$
      Hence $a_n = x_n-b_n$ is also convergent.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $$frac{1}{n^2} = b_n-b_{n-1}, ngeq2, b_1 = 1 $$
        then
        $$b_n = sum_{k=1}^n frac{1}{k^2} $$
        Then the following sequence is convergent, because it's increasing and bounded:
        $$x_n = b_n+a_n $$
        Hence $a_n = x_n-b_n$ is also convergent.






        share|cite|improve this answer









        $endgroup$



        $$frac{1}{n^2} = b_n-b_{n-1}, ngeq2, b_1 = 1 $$
        then
        $$b_n = sum_{k=1}^n frac{1}{k^2} $$
        Then the following sequence is convergent, because it's increasing and bounded:
        $$x_n = b_n+a_n $$
        Hence $a_n = x_n-b_n$ is also convergent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 13:37









        JakobianJakobian

        2,695721




        2,695721























            1












            $begingroup$

            First of all $a_{n}-a_{n+1}<frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
            For example the sequence $a_{n}=n$ satisfy the inequality.



            In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.



            The proof of your question is that: the series $1/n^{2}$ converges, so $forallvarepsilon>0,exists n_{varepsilon}$ tc$forall N>n_{varepsilon}$ $sum_{n=N}^{infty}1/n^{2}<varepsilon$.



            Because the sequence is upper bounded we can call $M:=limsup_{ntoinfty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $forallvarepsilon>0$ $exists k_{varepsilon}$ tc $|M-a_{n_{k_{varepsilon}}}|<varepsilon$.



            From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<sum_{i=n_{k}}^{n}1/i^{2}<sum_{i=n_{k}}^{infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-sum_{i=n_{k}}^{infty}1/i^{2}$.
            It follows that for $k>max{k_{varepsilon},n_{varepsilon}}$ $a_{n}>M-2varepsilon$ $forall n>n_k$.



            On the other hand let $M_{k}:=sup_{n>n_k}{a_{n}}$, so we have that obviously $a_{n}<M_{k}$ $forall n>k$, and it's clear because of the definition that $M_{k}to M$, and it does it decreasingly.
            So for $k>m_{varepsilon}$ we have $M_{k}<M+varepsilon$.



            So for $k>max{k_{varepsilon},n_{varepsilon}, m_{varepsilon}}$ and for $n>n_{k}$ we have $M+varepsilon>a_{n}>M-2varepsilon$, that means $a_nto M$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
              $endgroup$
              – Jakobian
              Dec 27 '18 at 15:21










            • $begingroup$
              You are right, i've changed my answer, hope it's right now
              $endgroup$
              – ecrin
              Dec 27 '18 at 16:55










            • $begingroup$
              I think it's ok
              $endgroup$
              – Jakobian
              Dec 27 '18 at 17:49
















            1












            $begingroup$

            First of all $a_{n}-a_{n+1}<frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
            For example the sequence $a_{n}=n$ satisfy the inequality.



            In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.



            The proof of your question is that: the series $1/n^{2}$ converges, so $forallvarepsilon>0,exists n_{varepsilon}$ tc$forall N>n_{varepsilon}$ $sum_{n=N}^{infty}1/n^{2}<varepsilon$.



            Because the sequence is upper bounded we can call $M:=limsup_{ntoinfty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $forallvarepsilon>0$ $exists k_{varepsilon}$ tc $|M-a_{n_{k_{varepsilon}}}|<varepsilon$.



            From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<sum_{i=n_{k}}^{n}1/i^{2}<sum_{i=n_{k}}^{infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-sum_{i=n_{k}}^{infty}1/i^{2}$.
            It follows that for $k>max{k_{varepsilon},n_{varepsilon}}$ $a_{n}>M-2varepsilon$ $forall n>n_k$.



            On the other hand let $M_{k}:=sup_{n>n_k}{a_{n}}$, so we have that obviously $a_{n}<M_{k}$ $forall n>k$, and it's clear because of the definition that $M_{k}to M$, and it does it decreasingly.
            So for $k>m_{varepsilon}$ we have $M_{k}<M+varepsilon$.



            So for $k>max{k_{varepsilon},n_{varepsilon}, m_{varepsilon}}$ and for $n>n_{k}$ we have $M+varepsilon>a_{n}>M-2varepsilon$, that means $a_nto M$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
              $endgroup$
              – Jakobian
              Dec 27 '18 at 15:21










            • $begingroup$
              You are right, i've changed my answer, hope it's right now
              $endgroup$
              – ecrin
              Dec 27 '18 at 16:55










            • $begingroup$
              I think it's ok
              $endgroup$
              – Jakobian
              Dec 27 '18 at 17:49














            1












            1








            1





            $begingroup$

            First of all $a_{n}-a_{n+1}<frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
            For example the sequence $a_{n}=n$ satisfy the inequality.



            In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.



            The proof of your question is that: the series $1/n^{2}$ converges, so $forallvarepsilon>0,exists n_{varepsilon}$ tc$forall N>n_{varepsilon}$ $sum_{n=N}^{infty}1/n^{2}<varepsilon$.



            Because the sequence is upper bounded we can call $M:=limsup_{ntoinfty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $forallvarepsilon>0$ $exists k_{varepsilon}$ tc $|M-a_{n_{k_{varepsilon}}}|<varepsilon$.



            From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<sum_{i=n_{k}}^{n}1/i^{2}<sum_{i=n_{k}}^{infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-sum_{i=n_{k}}^{infty}1/i^{2}$.
            It follows that for $k>max{k_{varepsilon},n_{varepsilon}}$ $a_{n}>M-2varepsilon$ $forall n>n_k$.



            On the other hand let $M_{k}:=sup_{n>n_k}{a_{n}}$, so we have that obviously $a_{n}<M_{k}$ $forall n>k$, and it's clear because of the definition that $M_{k}to M$, and it does it decreasingly.
            So for $k>m_{varepsilon}$ we have $M_{k}<M+varepsilon$.



            So for $k>max{k_{varepsilon},n_{varepsilon}, m_{varepsilon}}$ and for $n>n_{k}$ we have $M+varepsilon>a_{n}>M-2varepsilon$, that means $a_nto M$.






            share|cite|improve this answer











            $endgroup$



            First of all $a_{n}-a_{n+1}<frac{1}{n^{2}}$ doesn't mean the sequence is monotonically decreasing, on the contrary means that tends to be monotonically increasing.
            For example the sequence $a_{n}=n$ satisfy the inequality.



            In fact $a_{n}-a_{n+1}< 0$ means monotonically increasing and $a_{n}-a_{n+1}>0$ decreasing.



            The proof of your question is that: the series $1/n^{2}$ converges, so $forallvarepsilon>0,exists n_{varepsilon}$ tc$forall N>n_{varepsilon}$ $sum_{n=N}^{infty}1/n^{2}<varepsilon$.



            Because the sequence is upper bounded we can call $M:=limsup_{ntoinfty}a_{n}$, and we have a subsequence $a_{n_{k}}$ which converges to M. So $forallvarepsilon>0$ $exists k_{varepsilon}$ tc $|M-a_{n_{k_{varepsilon}}}|<varepsilon$.



            From the hypotesis we know that if $n>n_{k}$ then $a_{n_{k}}-a_{n}<sum_{i=n_{k}}^{n}1/i^{2}<sum_{i=n_{k}}^{infty}1/i^{2}$, so $a_{n}>a_{n_{k}}-sum_{i=n_{k}}^{infty}1/i^{2}$.
            It follows that for $k>max{k_{varepsilon},n_{varepsilon}}$ $a_{n}>M-2varepsilon$ $forall n>n_k$.



            On the other hand let $M_{k}:=sup_{n>n_k}{a_{n}}$, so we have that obviously $a_{n}<M_{k}$ $forall n>k$, and it's clear because of the definition that $M_{k}to M$, and it does it decreasingly.
            So for $k>m_{varepsilon}$ we have $M_{k}<M+varepsilon$.



            So for $k>max{k_{varepsilon},n_{varepsilon}, m_{varepsilon}}$ and for $n>n_{k}$ we have $M+varepsilon>a_{n}>M-2varepsilon$, that means $a_nto M$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 27 '18 at 17:18

























            answered Dec 27 '18 at 15:00









            ecrinecrin

            3477




            3477












            • $begingroup$
              Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
              $endgroup$
              – Jakobian
              Dec 27 '18 at 15:21










            • $begingroup$
              You are right, i've changed my answer, hope it's right now
              $endgroup$
              – ecrin
              Dec 27 '18 at 16:55










            • $begingroup$
              I think it's ok
              $endgroup$
              – Jakobian
              Dec 27 '18 at 17:49


















            • $begingroup$
              Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
              $endgroup$
              – Jakobian
              Dec 27 '18 at 15:21










            • $begingroup$
              You are right, i've changed my answer, hope it's right now
              $endgroup$
              – ecrin
              Dec 27 '18 at 16:55










            • $begingroup$
              I think it's ok
              $endgroup$
              – Jakobian
              Dec 27 '18 at 17:49
















            $begingroup$
            Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
            $endgroup$
            – Jakobian
            Dec 27 '18 at 15:21




            $begingroup$
            Where does $|a_{n_{k_varepsilon}}-a_n|<sum_{i=k_varepsilon}^n frac{1}{i^2}$ comes from?
            $endgroup$
            – Jakobian
            Dec 27 '18 at 15:21












            $begingroup$
            You are right, i've changed my answer, hope it's right now
            $endgroup$
            – ecrin
            Dec 27 '18 at 16:55




            $begingroup$
            You are right, i've changed my answer, hope it's right now
            $endgroup$
            – ecrin
            Dec 27 '18 at 16:55












            $begingroup$
            I think it's ok
            $endgroup$
            – Jakobian
            Dec 27 '18 at 17:49




            $begingroup$
            I think it's ok
            $endgroup$
            – Jakobian
            Dec 27 '18 at 17:49


















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