Prove that if $a_n$ is strictly decreasing, $a_n rightarrow 0$ and $n_0$ is even, then $ sum_{n=n_0}^{+ infty...












1












$begingroup$


I have a problem with this task because I don't know how much I must write here.
If $a_n$ is strictly decreasing and $a_n rightarrow 0$, then I know that $a_n$ is limited from below by $0$ and all words of $a_n$ are $>0$, because if $a_n<0$ and $a_n$ strictly decreasing, then $a_n rightarrow - infty$, but it is contradiction.
If I know that all $a_n$ is $>0$ (because I prove it), then I know that $a_{n_0}+a_{n_1}+a_{n_2}+...= sum_{n=n_0}^{+ infty } a_n>0$. However I am afraid that this is not enough evidence. Moreover I don't know for what purpose the information is given that $ a_ {n_0} $ is even.






real-analysis







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  • 1




    $begingroup$
    The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
    $endgroup$
    – Christoph
    Dec 20 '18 at 11:20
















1












$begingroup$


I have a problem with this task because I don't know how much I must write here.
If $a_n$ is strictly decreasing and $a_n rightarrow 0$, then I know that $a_n$ is limited from below by $0$ and all words of $a_n$ are $>0$, because if $a_n<0$ and $a_n$ strictly decreasing, then $a_n rightarrow - infty$, but it is contradiction.
If I know that all $a_n$ is $>0$ (because I prove it), then I know that $a_{n_0}+a_{n_1}+a_{n_2}+...= sum_{n=n_0}^{+ infty } a_n>0$. However I am afraid that this is not enough evidence. Moreover I don't know for what purpose the information is given that $ a_ {n_0} $ is even.






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
    $endgroup$
    – Christoph
    Dec 20 '18 at 11:20














1












1








1





$begingroup$


I have a problem with this task because I don't know how much I must write here.
If $a_n$ is strictly decreasing and $a_n rightarrow 0$, then I know that $a_n$ is limited from below by $0$ and all words of $a_n$ are $>0$, because if $a_n<0$ and $a_n$ strictly decreasing, then $a_n rightarrow - infty$, but it is contradiction.
If I know that all $a_n$ is $>0$ (because I prove it), then I know that $a_{n_0}+a_{n_1}+a_{n_2}+...= sum_{n=n_0}^{+ infty } a_n>0$. However I am afraid that this is not enough evidence. Moreover I don't know for what purpose the information is given that $ a_ {n_0} $ is even.






real-analysis







share|cite|improve this question









$endgroup$




I have a problem with this task because I don't know how much I must write here.
If $a_n$ is strictly decreasing and $a_n rightarrow 0$, then I know that $a_n$ is limited from below by $0$ and all words of $a_n$ are $>0$, because if $a_n<0$ and $a_n$ strictly decreasing, then $a_n rightarrow - infty$, but it is contradiction.
If I know that all $a_n$ is $>0$ (because I prove it), then I know that $a_{n_0}+a_{n_1}+a_{n_2}+...= sum_{n=n_0}^{+ infty } a_n>0$. However I am afraid that this is not enough evidence. Moreover I don't know for what purpose the information is given that $ a_ {n_0} $ is even.






real-analysis




real-analysis






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asked Dec 20 '18 at 10:48









MP3129MP3129

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  • 1




    $begingroup$
    The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
    $endgroup$
    – Christoph
    Dec 20 '18 at 11:20














  • 1




    $begingroup$
    The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
    $endgroup$
    – Christoph
    Dec 20 '18 at 11:20








1




1




$begingroup$
The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
$endgroup$
– Christoph
Dec 20 '18 at 11:20




$begingroup$
The information that $n_0$ (or $a_{n_0}$ as in the body of the question?) is even is irrelevant. Did you cite the problem statement correctly?
$endgroup$
– Christoph
Dec 20 '18 at 11:20










1 Answer
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$begingroup$

Your assumption that if $exists nin mathbf{N}$ such that $a_n<0$ then that ${a_n}$ is strictly decreasing should imply that $a_nrightarrow -infty$ is false. Consider $a_n = -1+frac{1}{n}$ which is strictly decreasing but $a_nrightarrow -1$ as $nrightarrow infty$.



Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_nnotrightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.



The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then



$$sum_{n=n_0}^{infty}a_n>0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
    $endgroup$
    – MP3129
    Dec 20 '18 at 11:44











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1 Answer
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1 Answer
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active

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active

oldest

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3












$begingroup$

Your assumption that if $exists nin mathbf{N}$ such that $a_n<0$ then that ${a_n}$ is strictly decreasing should imply that $a_nrightarrow -infty$ is false. Consider $a_n = -1+frac{1}{n}$ which is strictly decreasing but $a_nrightarrow -1$ as $nrightarrow infty$.



Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_nnotrightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.



The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then



$$sum_{n=n_0}^{infty}a_n>0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
    $endgroup$
    – MP3129
    Dec 20 '18 at 11:44
















3












$begingroup$

Your assumption that if $exists nin mathbf{N}$ such that $a_n<0$ then that ${a_n}$ is strictly decreasing should imply that $a_nrightarrow -infty$ is false. Consider $a_n = -1+frac{1}{n}$ which is strictly decreasing but $a_nrightarrow -1$ as $nrightarrow infty$.



Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_nnotrightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.



The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then



$$sum_{n=n_0}^{infty}a_n>0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
    $endgroup$
    – MP3129
    Dec 20 '18 at 11:44














3












3








3





$begingroup$

Your assumption that if $exists nin mathbf{N}$ such that $a_n<0$ then that ${a_n}$ is strictly decreasing should imply that $a_nrightarrow -infty$ is false. Consider $a_n = -1+frac{1}{n}$ which is strictly decreasing but $a_nrightarrow -1$ as $nrightarrow infty$.



Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_nnotrightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.



The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then



$$sum_{n=n_0}^{infty}a_n>0$$






share|cite|improve this answer









$endgroup$



Your assumption that if $exists nin mathbf{N}$ such that $a_n<0$ then that ${a_n}$ is strictly decreasing should imply that $a_nrightarrow -infty$ is false. Consider $a_n = -1+frac{1}{n}$ which is strictly decreasing but $a_nrightarrow -1$ as $nrightarrow infty$.



Instead if $a_{n_0}<0$ for some $n_{0}$ then you can conclude that $a_nnotrightarrow 0$. Do you see how? Consider some neighborhood of $0$ which does not contain any $a_n$ with $n>n_0$.



The fact that $n_0$ is even is immaterial. If $a_n>0$ for every $n$ then



$$sum_{n=n_0}^{infty}a_n>0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 11:06









Olof RubinOlof Rubin

1,141316




1,141316












  • $begingroup$
    Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
    $endgroup$
    – MP3129
    Dec 20 '18 at 11:44


















  • $begingroup$
    Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
    $endgroup$
    – MP3129
    Dec 20 '18 at 11:44
















$begingroup$
Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
$endgroup$
– MP3129
Dec 20 '18 at 11:44




$begingroup$
Yes, you have right that I had a mistake in reasoning regarding $a_{n_0}<0$. Thank's for help
$endgroup$
– MP3129
Dec 20 '18 at 11:44


















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