If $lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) =[liminf(a_n),limsup(a_n)]$












0















If ${a_n}$ is a series, and $lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) = [liminf(a_n),limsup(a_n)]$




Basically, I wish to prove that the set of all partial limits of the series $a_n$ (as $ntoinfty$) contains all values between its maximal limit ($limsup$) and minimal limit ($liminf$)



I already know that $limsup$ and $liminf$ are limits of $a_n$ by definition, and so it is a closed set $[liminf(a_n),limsup(a_n)]$.



Since $limsup(a_n) = sup(P(a_n))$ and $liminf(a_n) = inf(P(a_n))$ then $P(a_n)$ is included in $[liminf(a_n),limsup(a_n)]$.



Now if I were to take any $liminf(a_n) < L < limsup(a_n)$, I need to prove that it is a partial limit.






sequences-and-series limsup-and-liminf







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    0















    If ${a_n}$ is a series, and $lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) = [liminf(a_n),limsup(a_n)]$




    Basically, I wish to prove that the set of all partial limits of the series $a_n$ (as $ntoinfty$) contains all values between its maximal limit ($limsup$) and minimal limit ($liminf$)



    I already know that $limsup$ and $liminf$ are limits of $a_n$ by definition, and so it is a closed set $[liminf(a_n),limsup(a_n)]$.



    Since $limsup(a_n) = sup(P(a_n))$ and $liminf(a_n) = inf(P(a_n))$ then $P(a_n)$ is included in $[liminf(a_n),limsup(a_n)]$.



    Now if I were to take any $liminf(a_n) < L < limsup(a_n)$, I need to prove that it is a partial limit.






    sequences-and-series limsup-and-liminf







    share|cite|improve this question



























      0












      0








      0


      1






      If ${a_n}$ is a series, and $lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) = [liminf(a_n),limsup(a_n)]$




      Basically, I wish to prove that the set of all partial limits of the series $a_n$ (as $ntoinfty$) contains all values between its maximal limit ($limsup$) and minimal limit ($liminf$)



      I already know that $limsup$ and $liminf$ are limits of $a_n$ by definition, and so it is a closed set $[liminf(a_n),limsup(a_n)]$.



      Since $limsup(a_n) = sup(P(a_n))$ and $liminf(a_n) = inf(P(a_n))$ then $P(a_n)$ is included in $[liminf(a_n),limsup(a_n)]$.



      Now if I were to take any $liminf(a_n) < L < limsup(a_n)$, I need to prove that it is a partial limit.






      sequences-and-series limsup-and-liminf







      share|cite|improve this question
















      If ${a_n}$ is a series, and $lim(a_{n+1} - a_n) = 0$, prove that $P(a_n) = [liminf(a_n),limsup(a_n)]$




      Basically, I wish to prove that the set of all partial limits of the series $a_n$ (as $ntoinfty$) contains all values between its maximal limit ($limsup$) and minimal limit ($liminf$)



      I already know that $limsup$ and $liminf$ are limits of $a_n$ by definition, and so it is a closed set $[liminf(a_n),limsup(a_n)]$.



      Since $limsup(a_n) = sup(P(a_n))$ and $liminf(a_n) = inf(P(a_n))$ then $P(a_n)$ is included in $[liminf(a_n),limsup(a_n)]$.



      Now if I were to take any $liminf(a_n) < L < limsup(a_n)$, I need to prove that it is a partial limit.






      sequences-and-series limsup-and-liminf




      sequences-and-series limsup-and-liminf






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      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 25 '18 at 16:22









      Tianlalu

      3,09621038




      3,09621038










      asked Nov 25 '18 at 16:12









      Boaz Yakubov

      254




      254






















          1 Answer
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          Let $liminf_n a_n < L < limsup_n a_n$, and let $k_0inmathbb{Z}^+$ be such that
          $$
          liminf_n a_n < L-frac{1}{k_0},,
          qquad
          L + frac{1}{k_0} < limsup_n a_n.
          $$
          By assumption, there exists an index $N_0$ such that
          $$
          (1)qquad
          |a_{n+1}-a_n| < frac{1}{k_0}
          qquad forall ngeq N_0.
          $$
          Moreover, we can find indices $j_0, m_0 geq N_0$ such that
          $$
          a_{j_0} < L - frac{1}{k_0},
          qquad
          L + frac{1}{k_0} < a_{m_0}.
          $$
          Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
          $n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < frac{1}{k_0}$.



          As a second step, let $N_1 > max{N_0, j_0, m_0}$ be such that
          $$
          (2)qquad
          |a_{n+1}-a_n| < frac{1}{k_0+1}
          qquad forall ngeq N_1.
          $$
          Moreover, we can find indices $j_1, m_1 geq N_1$ such that
          $$
          a_{j_1} < L - frac{1}{k_0+1},
          qquad
          L + frac{1}{k_0+1} < a_{m_1}.
          $$
          Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
          $n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < frac{1}{k_0+1}$.



          We can in this way construct by induction a sequence $n_0 < n_1 < cdots$ such that
          $$
          |a_{n_j} - L| < frac{1}{k_0+j}
          qquadforall j,
          $$
          i.e. $lim_j a_{n_j} = L$.






          share|cite|improve this answer





















          • Marvelous, thanks!
            – Boaz Yakubov
            Nov 25 '18 at 18:10











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          1














          Let $liminf_n a_n < L < limsup_n a_n$, and let $k_0inmathbb{Z}^+$ be such that
          $$
          liminf_n a_n < L-frac{1}{k_0},,
          qquad
          L + frac{1}{k_0} < limsup_n a_n.
          $$
          By assumption, there exists an index $N_0$ such that
          $$
          (1)qquad
          |a_{n+1}-a_n| < frac{1}{k_0}
          qquad forall ngeq N_0.
          $$
          Moreover, we can find indices $j_0, m_0 geq N_0$ such that
          $$
          a_{j_0} < L - frac{1}{k_0},
          qquad
          L + frac{1}{k_0} < a_{m_0}.
          $$
          Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
          $n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < frac{1}{k_0}$.



          As a second step, let $N_1 > max{N_0, j_0, m_0}$ be such that
          $$
          (2)qquad
          |a_{n+1}-a_n| < frac{1}{k_0+1}
          qquad forall ngeq N_1.
          $$
          Moreover, we can find indices $j_1, m_1 geq N_1$ such that
          $$
          a_{j_1} < L - frac{1}{k_0+1},
          qquad
          L + frac{1}{k_0+1} < a_{m_1}.
          $$
          Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
          $n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < frac{1}{k_0+1}$.



          We can in this way construct by induction a sequence $n_0 < n_1 < cdots$ such that
          $$
          |a_{n_j} - L| < frac{1}{k_0+j}
          qquadforall j,
          $$
          i.e. $lim_j a_{n_j} = L$.






          share|cite|improve this answer





















          • Marvelous, thanks!
            – Boaz Yakubov
            Nov 25 '18 at 18:10
















          1














          Let $liminf_n a_n < L < limsup_n a_n$, and let $k_0inmathbb{Z}^+$ be such that
          $$
          liminf_n a_n < L-frac{1}{k_0},,
          qquad
          L + frac{1}{k_0} < limsup_n a_n.
          $$
          By assumption, there exists an index $N_0$ such that
          $$
          (1)qquad
          |a_{n+1}-a_n| < frac{1}{k_0}
          qquad forall ngeq N_0.
          $$
          Moreover, we can find indices $j_0, m_0 geq N_0$ such that
          $$
          a_{j_0} < L - frac{1}{k_0},
          qquad
          L + frac{1}{k_0} < a_{m_0}.
          $$
          Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
          $n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < frac{1}{k_0}$.



          As a second step, let $N_1 > max{N_0, j_0, m_0}$ be such that
          $$
          (2)qquad
          |a_{n+1}-a_n| < frac{1}{k_0+1}
          qquad forall ngeq N_1.
          $$
          Moreover, we can find indices $j_1, m_1 geq N_1$ such that
          $$
          a_{j_1} < L - frac{1}{k_0+1},
          qquad
          L + frac{1}{k_0+1} < a_{m_1}.
          $$
          Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
          $n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < frac{1}{k_0+1}$.



          We can in this way construct by induction a sequence $n_0 < n_1 < cdots$ such that
          $$
          |a_{n_j} - L| < frac{1}{k_0+j}
          qquadforall j,
          $$
          i.e. $lim_j a_{n_j} = L$.






          share|cite|improve this answer





















          • Marvelous, thanks!
            – Boaz Yakubov
            Nov 25 '18 at 18:10














          1












          1








          1






          Let $liminf_n a_n < L < limsup_n a_n$, and let $k_0inmathbb{Z}^+$ be such that
          $$
          liminf_n a_n < L-frac{1}{k_0},,
          qquad
          L + frac{1}{k_0} < limsup_n a_n.
          $$
          By assumption, there exists an index $N_0$ such that
          $$
          (1)qquad
          |a_{n+1}-a_n| < frac{1}{k_0}
          qquad forall ngeq N_0.
          $$
          Moreover, we can find indices $j_0, m_0 geq N_0$ such that
          $$
          a_{j_0} < L - frac{1}{k_0},
          qquad
          L + frac{1}{k_0} < a_{m_0}.
          $$
          Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
          $n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < frac{1}{k_0}$.



          As a second step, let $N_1 > max{N_0, j_0, m_0}$ be such that
          $$
          (2)qquad
          |a_{n+1}-a_n| < frac{1}{k_0+1}
          qquad forall ngeq N_1.
          $$
          Moreover, we can find indices $j_1, m_1 geq N_1$ such that
          $$
          a_{j_1} < L - frac{1}{k_0+1},
          qquad
          L + frac{1}{k_0+1} < a_{m_1}.
          $$
          Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
          $n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < frac{1}{k_0+1}$.



          We can in this way construct by induction a sequence $n_0 < n_1 < cdots$ such that
          $$
          |a_{n_j} - L| < frac{1}{k_0+j}
          qquadforall j,
          $$
          i.e. $lim_j a_{n_j} = L$.






          share|cite|improve this answer












          Let $liminf_n a_n < L < limsup_n a_n$, and let $k_0inmathbb{Z}^+$ be such that
          $$
          liminf_n a_n < L-frac{1}{k_0},,
          qquad
          L + frac{1}{k_0} < limsup_n a_n.
          $$
          By assumption, there exists an index $N_0$ such that
          $$
          (1)qquad
          |a_{n+1}-a_n| < frac{1}{k_0}
          qquad forall ngeq N_0.
          $$
          Moreover, we can find indices $j_0, m_0 geq N_0$ such that
          $$
          a_{j_0} < L - frac{1}{k_0},
          qquad
          L + frac{1}{k_0} < a_{m_0}.
          $$
          Assume for example that $j_0 < m_0$. Then, by (1), there exists an index
          $n_0$ such that $j_0 < n_0 < m_0$ and $|a_{n_0} - L| < frac{1}{k_0}$.



          As a second step, let $N_1 > max{N_0, j_0, m_0}$ be such that
          $$
          (2)qquad
          |a_{n+1}-a_n| < frac{1}{k_0+1}
          qquad forall ngeq N_1.
          $$
          Moreover, we can find indices $j_1, m_1 geq N_1$ such that
          $$
          a_{j_1} < L - frac{1}{k_0+1},
          qquad
          L + frac{1}{k_0+1} < a_{m_1}.
          $$
          Assume for example that $j_1 < m_1$. Then, by (2), there exists an index
          $n_1$ such that $j_1 < n_1 < m_1$ and $|a_{n_1} - L| < frac{1}{k_0+1}$.



          We can in this way construct by induction a sequence $n_0 < n_1 < cdots$ such that
          $$
          |a_{n_j} - L| < frac{1}{k_0+j}
          qquadforall j,
          $$
          i.e. $lim_j a_{n_j} = L$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 17:52









          Rigel

          10.9k11320




          10.9k11320












          • Marvelous, thanks!
            – Boaz Yakubov
            Nov 25 '18 at 18:10


















          • Marvelous, thanks!
            – Boaz Yakubov
            Nov 25 '18 at 18:10
















          Marvelous, thanks!
          – Boaz Yakubov
          Nov 25 '18 at 18:10




          Marvelous, thanks!
          – Boaz Yakubov
          Nov 25 '18 at 18:10


















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