Show that the following sequence converges. Please Critique my proof.












9












$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$

Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$

and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.










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$endgroup$








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    8 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    7 hours ago


















9












$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$

Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$

and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    8 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    7 hours ago
















9












9








9


1



$begingroup$


The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$

Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$

and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.










share|cite|improve this question











$endgroup$




The problem is as follows:




Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$

Show that $a_n$ converges.




My (wrong) proof:



Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$

and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.



My question:
This is a question from a comprehensive exam I found and am using to review.



Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?



Notes: Currently working on the proof.







real-analysis sequences-and-series convergence fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







Darel

















asked 8 hours ago









DarelDarel

1199




1199








  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    8 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    7 hours ago
















  • 4




    $begingroup$
    Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
    $endgroup$
    – Falrach
    8 hours ago






  • 1




    $begingroup$
    Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
    $endgroup$
    – Maximilian Janisch
    7 hours ago










4




4




$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
8 hours ago




$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
8 hours ago




1




1




$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
7 hours ago






$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
7 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$



which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    6 hours ago





















1












$begingroup$

Define $b_k := a_{2k+1}$. Then
$$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k to b$.
Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
begin{align}
a_{2m+1} - a_m < - frac{varepsilon}{2}
end{align}

for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
begin{align*}
d_m := 1_M (m)
end{align*}

This implies
begin{align*}
0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
end{align*}

since $|M| = infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
begin{align*}
|a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
end{align*}

for infinitely $n geq K$. Contradiction. Thus $a_n to b$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



    $$ b_{n+1}
    = a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    = b_n, $$



    which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
      $endgroup$
      – Mars Plastic
      6 hours ago


















    3












    $begingroup$

    Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



    $$ b_{n+1}
    = a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    = b_n, $$



    which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
      $endgroup$
      – Mars Plastic
      6 hours ago
















    3












    3








    3





    $begingroup$

    Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



    $$ b_{n+1}
    = a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    = b_n, $$



    which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.






    share|cite|improve this answer









    $endgroup$



    Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then



    $$ b_{n+1}
    = a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
    = b_n, $$



    which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    Sangchul LeeSangchul Lee

    95k12170276




    95k12170276








    • 2




      $begingroup$
      Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
      $endgroup$
      – Mars Plastic
      6 hours ago
















    • 2




      $begingroup$
      Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
      $endgroup$
      – Mars Plastic
      6 hours ago










    2




    2




    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    6 hours ago






    $begingroup$
    Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
    $endgroup$
    – Mars Plastic
    6 hours ago













    1












    $begingroup$

    Define $b_k := a_{2k+1}$. Then
    $$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
    Since $b_k$ is non-negative and non-increasing: $b_k to b$.
    Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
    Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
    begin{align}
    a_{2m+1} - a_m < - frac{varepsilon}{2}
    end{align}

    for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
    begin{align*}
    d_m := 1_M (m)
    end{align*}

    This implies
    begin{align*}
    0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
    leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
    end{align*}

    since $|M| = infty$ and the last series converges. This is a contradiction.
    Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
    begin{align*}
    |a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
    end{align*}

    for infinitely $n geq K$. Contradiction. Thus $a_n to b$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Define $b_k := a_{2k+1}$. Then
      $$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
      Since $b_k$ is non-negative and non-increasing: $b_k to b$.
      Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
      Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
      begin{align}
      a_{2m+1} - a_m < - frac{varepsilon}{2}
      end{align}

      for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
      begin{align*}
      d_m := 1_M (m)
      end{align*}

      This implies
      begin{align*}
      0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
      leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
      end{align*}

      since $|M| = infty$ and the last series converges. This is a contradiction.
      Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
      begin{align*}
      |a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
      end{align*}

      for infinitely $n geq K$. Contradiction. Thus $a_n to b$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Define $b_k := a_{2k+1}$. Then
        $$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
        Since $b_k$ is non-negative and non-increasing: $b_k to b$.
        Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
        Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
        begin{align}
        a_{2m+1} - a_m < - frac{varepsilon}{2}
        end{align}

        for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
        begin{align*}
        d_m := 1_M (m)
        end{align*}

        This implies
        begin{align*}
        0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
        leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
        end{align*}

        since $|M| = infty$ and the last series converges. This is a contradiction.
        Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
        begin{align*}
        |a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
        end{align*}

        for infinitely $n geq K$. Contradiction. Thus $a_n to b$.






        share|cite|improve this answer









        $endgroup$



        Define $b_k := a_{2k+1}$. Then
        $$b_k leq a_{2k} + (-1)^{2k}frac{1}{2k} leq b_{k-1} + (frac{1}{2k} - frac{1}{2k-1}) leq b_{k-1}$$
        Since $b_k$ is non-negative and non-increasing: $b_k to b$.
        Suppose $a_n nrightarrow b$. Then there exists an $varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > varepsilon$.
        Assume that $|a_{2m+1}-a_m| > frac{varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m leq frac{1}{2m}$ we have that
        begin{align}
        a_{2m+1} - a_m < - frac{varepsilon}{2}
        end{align}

        for infinitely many $m$. Let $M := {m geq 1 : a_{2m+1} - a_m < - frac{varepsilon}{2} text{ is fulfilled for } m }$
        begin{align*}
        d_m := 1_M (m)
        end{align*}

        This implies
        begin{align*}
        0 leq a_{2m+1} = a_1 + sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + sum_{k=1}^m (a_{2k+1} - a_{2k}) + sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \
        leq a_1 + sum_{k=1}^m (-1)^{2k} frac{1}{2k}- frac{varepsilon}{2} d_k + sum_{k=1}^m (-1)^{2k-1}frac{1}{2k-1} to a_1 - sum_{k=1}^infty frac{varepsilon}{2} d_k + sum_{i=1}^infty (-1)^i frac{1}{i} = - infty
        end{align*}

        since $|M| = infty$ and the last series converges. This is a contradiction.
        Therefore we have that there exists $Kgeq 1$ s.t. for all $kgeq K$ it holds: $|a_{2k+1} - a_k| leq frac{varepsilon}{2}$. We can conclude that
        begin{align*}
        |a{2n+1} - b| geq |a_{2n} - b| - |a_{2n+1} - a_n| geq varepsilon - frac{varepsilon}{2} = frac{varepsilon}{2}
        end{align*}

        for infinitely $n geq K$. Contradiction. Thus $a_n to b$.







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        answered 4 hours ago









        FalrachFalrach

        1,676224




        1,676224






























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