Given $lim_{ntoinfty}x_n = infty$ show that $y_n = left{ sum_{k=1}^n x_kright}$ is an unbounded sequence.












0












$begingroup$



Given:
$$
lim_{ntoinfty}x_n = infty
$$
show that
$$
y_n = left{ sum_{k=1}^n x_kright}
$$
is an unbounded sequence.




Intuitively this is obvious, however I'm having a hard time proving that formally.



Start with definition of a divergent sequence we have that for any $varepsilon > 0$ there exists some $N$ such that $x_n > varepsilon$:
$$
forall varepsilon >0 exists NinBbb N: forall n > N implies x_n > varepsilon
$$



With that in mind there must be some index starting from which the sequence becomes a monotonically increasing sequence. Now if we consider the difference between sums we may obtain:
$$
y_{n} - y_{n-1} = x_n \
y_{n+1} - y_n = x_{n+1}
$$



So from this we (hopefully) may conclude that the difference between the terms of $y_n$ is also increasing which means that the whole sum is also monotonically increasing which means it has no upper bound.



The problem with the above is that it doesn't feel like a formal proof and I have strong doubts about the validity of my reasoning.



Eventually the question is how to prove what's in the problem section in valid formal way?






calculus sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    $lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
    $endgroup$
    – Arthur
    Dec 9 '18 at 19:59










  • $begingroup$
    @Arthur I've made an edit, thank you for the notice
    $endgroup$
    – roman
    Dec 9 '18 at 20:02










  • $begingroup$
    In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
    $endgroup$
    – RRL
    Dec 9 '18 at 20:03


















0












$begingroup$



Given:
$$
lim_{ntoinfty}x_n = infty
$$
show that
$$
y_n = left{ sum_{k=1}^n x_kright}
$$
is an unbounded sequence.




Intuitively this is obvious, however I'm having a hard time proving that formally.



Start with definition of a divergent sequence we have that for any $varepsilon > 0$ there exists some $N$ such that $x_n > varepsilon$:
$$
forall varepsilon >0 exists NinBbb N: forall n > N implies x_n > varepsilon
$$



With that in mind there must be some index starting from which the sequence becomes a monotonically increasing sequence. Now if we consider the difference between sums we may obtain:
$$
y_{n} - y_{n-1} = x_n \
y_{n+1} - y_n = x_{n+1}
$$



So from this we (hopefully) may conclude that the difference between the terms of $y_n$ is also increasing which means that the whole sum is also monotonically increasing which means it has no upper bound.



The problem with the above is that it doesn't feel like a formal proof and I have strong doubts about the validity of my reasoning.



Eventually the question is how to prove what's in the problem section in valid formal way?






calculus sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    $lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
    $endgroup$
    – Arthur
    Dec 9 '18 at 19:59










  • $begingroup$
    @Arthur I've made an edit, thank you for the notice
    $endgroup$
    – roman
    Dec 9 '18 at 20:02










  • $begingroup$
    In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
    $endgroup$
    – RRL
    Dec 9 '18 at 20:03
















0












0








0





$begingroup$



Given:
$$
lim_{ntoinfty}x_n = infty
$$
show that
$$
y_n = left{ sum_{k=1}^n x_kright}
$$
is an unbounded sequence.




Intuitively this is obvious, however I'm having a hard time proving that formally.



Start with definition of a divergent sequence we have that for any $varepsilon > 0$ there exists some $N$ such that $x_n > varepsilon$:
$$
forall varepsilon >0 exists NinBbb N: forall n > N implies x_n > varepsilon
$$



With that in mind there must be some index starting from which the sequence becomes a monotonically increasing sequence. Now if we consider the difference between sums we may obtain:
$$
y_{n} - y_{n-1} = x_n \
y_{n+1} - y_n = x_{n+1}
$$



So from this we (hopefully) may conclude that the difference between the terms of $y_n$ is also increasing which means that the whole sum is also monotonically increasing which means it has no upper bound.



The problem with the above is that it doesn't feel like a formal proof and I have strong doubts about the validity of my reasoning.



Eventually the question is how to prove what's in the problem section in valid formal way?






calculus sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$





Given:
$$
lim_{ntoinfty}x_n = infty
$$
show that
$$
y_n = left{ sum_{k=1}^n x_kright}
$$
is an unbounded sequence.




Intuitively this is obvious, however I'm having a hard time proving that formally.



Start with definition of a divergent sequence we have that for any $varepsilon > 0$ there exists some $N$ such that $x_n > varepsilon$:
$$
forall varepsilon >0 exists NinBbb N: forall n > N implies x_n > varepsilon
$$



With that in mind there must be some index starting from which the sequence becomes a monotonically increasing sequence. Now if we consider the difference between sums we may obtain:
$$
y_{n} - y_{n-1} = x_n \
y_{n+1} - y_n = x_{n+1}
$$



So from this we (hopefully) may conclude that the difference between the terms of $y_n$ is also increasing which means that the whole sum is also monotonically increasing which means it has no upper bound.



The problem with the above is that it doesn't feel like a formal proof and I have strong doubts about the validity of my reasoning.



Eventually the question is how to prove what's in the problem section in valid formal way?






calculus sequences-and-series limits proof-verification




calculus sequences-and-series limits proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 20:02







roman

















asked Dec 9 '18 at 19:52









romanroman

2,16321223




2,16321223












  • $begingroup$
    $lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
    $endgroup$
    – Arthur
    Dec 9 '18 at 19:59










  • $begingroup$
    @Arthur I've made an edit, thank you for the notice
    $endgroup$
    – roman
    Dec 9 '18 at 20:02










  • $begingroup$
    In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
    $endgroup$
    – RRL
    Dec 9 '18 at 20:03




















  • $begingroup$
    $lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
    $endgroup$
    – Arthur
    Dec 9 '18 at 19:59










  • $begingroup$
    @Arthur I've made an edit, thank you for the notice
    $endgroup$
    – roman
    Dec 9 '18 at 20:02










  • $begingroup$
    In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
    $endgroup$
    – RRL
    Dec 9 '18 at 20:03


















$begingroup$
$lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
$endgroup$
– Arthur
Dec 9 '18 at 19:59




$begingroup$
$lim_{ntoinfty}x_n = infty$ specifically means $+infty$, not $pminfty$.
$endgroup$
– Arthur
Dec 9 '18 at 19:59












$begingroup$
@Arthur I've made an edit, thank you for the notice
$endgroup$
– roman
Dec 9 '18 at 20:02




$begingroup$
@Arthur I've made an edit, thank you for the notice
$endgroup$
– roman
Dec 9 '18 at 20:02












$begingroup$
In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
$endgroup$
– RRL
Dec 9 '18 at 20:03






$begingroup$
In the first case $sum_{k=N+1}^n x_k > epsilon(n -N) to +infty$ as $n to infty$ and the sum for $1 leqslant k leqslant N$ is fixed.
$endgroup$
– RRL
Dec 9 '18 at 20:03












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$begingroup$

So you have $x_n>epsilon>0$ for all $ngeq N.$ Now choose $Pinmathbb{R}$. By Archimedes choose $minmathbb{N}$ such that $mepsilon>P+|sum_{k=1}^Nx_n|$. Then we have
$$sum_{k=1}^{N+m}x_kgeq sum_{k=1}^{N}x_k+mepsilongeq-left|sum_{k=1}^{N}x_kright|+mepsilon>-left|sum_{k=1}^Nx_nright|+left(P+left|sum_{k=1}^Nx_nright|right)=P.$$






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    $begingroup$

    So you have $x_n>epsilon>0$ for all $ngeq N.$ Now choose $Pinmathbb{R}$. By Archimedes choose $minmathbb{N}$ such that $mepsilon>P+|sum_{k=1}^Nx_n|$. Then we have
    $$sum_{k=1}^{N+m}x_kgeq sum_{k=1}^{N}x_k+mepsilongeq-left|sum_{k=1}^{N}x_kright|+mepsilon>-left|sum_{k=1}^Nx_nright|+left(P+left|sum_{k=1}^Nx_nright|right)=P.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      So you have $x_n>epsilon>0$ for all $ngeq N.$ Now choose $Pinmathbb{R}$. By Archimedes choose $minmathbb{N}$ such that $mepsilon>P+|sum_{k=1}^Nx_n|$. Then we have
      $$sum_{k=1}^{N+m}x_kgeq sum_{k=1}^{N}x_k+mepsilongeq-left|sum_{k=1}^{N}x_kright|+mepsilon>-left|sum_{k=1}^Nx_nright|+left(P+left|sum_{k=1}^Nx_nright|right)=P.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        So you have $x_n>epsilon>0$ for all $ngeq N.$ Now choose $Pinmathbb{R}$. By Archimedes choose $minmathbb{N}$ such that $mepsilon>P+|sum_{k=1}^Nx_n|$. Then we have
        $$sum_{k=1}^{N+m}x_kgeq sum_{k=1}^{N}x_k+mepsilongeq-left|sum_{k=1}^{N}x_kright|+mepsilon>-left|sum_{k=1}^Nx_nright|+left(P+left|sum_{k=1}^Nx_nright|right)=P.$$






        share|cite|improve this answer









        $endgroup$



        So you have $x_n>epsilon>0$ for all $ngeq N.$ Now choose $Pinmathbb{R}$. By Archimedes choose $minmathbb{N}$ such that $mepsilon>P+|sum_{k=1}^Nx_n|$. Then we have
        $$sum_{k=1}^{N+m}x_kgeq sum_{k=1}^{N}x_k+mepsilongeq-left|sum_{k=1}^{N}x_kright|+mepsilon>-left|sum_{k=1}^Nx_nright|+left(P+left|sum_{k=1}^Nx_nright|right)=P.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 19:59









        MelodyMelody

        78812




        78812






























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