Prove that $lim_{n to infty} a_n = -infty $ if $ forall ngeq m: a_n leq a_m +2017$, and $-infty$ is a...
$begingroup$
Given a sequence ${a_n}_{n=1}^{infty} $ such that
$forall ngeq m: a_n leq a_m +2017$,
and $-infty$ is a substantial limit of $a_n$,
Prove that $lim_{n to infty} a_n = -infty$.
my attempt:
It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.
I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.
there exist a sub-sequence of $a_n$,
let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$
then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $
$exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $
I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.
real-analysis calculus sequences-and-series convergence
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
$endgroup$
add a comment |
$begingroup$
Given a sequence ${a_n}_{n=1}^{infty} $ such that
$forall ngeq m: a_n leq a_m +2017$,
and $-infty$ is a substantial limit of $a_n$,
Prove that $lim_{n to infty} a_n = -infty$.
my attempt:
It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.
I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.
there exist a sub-sequence of $a_n$,
let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$
then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $
$exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $
I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.
real-analysis calculus sequences-and-series convergence
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
$endgroup$
add a comment |
$begingroup$
Given a sequence ${a_n}_{n=1}^{infty} $ such that
$forall ngeq m: a_n leq a_m +2017$,
and $-infty$ is a substantial limit of $a_n$,
Prove that $lim_{n to infty} a_n = -infty$.
my attempt:
It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.
I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.
there exist a sub-sequence of $a_n$,
let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$
then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $
$exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $
I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.
real-analysis calculus sequences-and-series convergence
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
$endgroup$
Given a sequence ${a_n}_{n=1}^{infty} $ such that
$forall ngeq m: a_n leq a_m +2017$,
and $-infty$ is a substantial limit of $a_n$,
Prove that $lim_{n to infty} a_n = -infty$.
my attempt:
It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.
I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.
there exist a sub-sequence of $a_n$,
let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$
then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $
$exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $
I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.
real-analysis calculus sequences-and-series convergence
real-analysis calculus sequences-and-series convergence
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
edited Dec 5 '18 at 17:14
Martin Sleziak
44.8k9118272
44.8k9118272
asked Dec 5 '18 at 16:12
JnevenJneven
826322
asked Dec 5 '18 at 16:12
JnevenJneven
826322
asked Dec 5 '18 at 16:12
JnevenJneven
826322
826322
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
$endgroup$
add a comment |
$begingroup$
If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
$endgroup$
add a comment |
$begingroup$
Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
$endgroup$
add a comment |
$begingroup$
Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
$endgroup$
Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
answered Dec 5 '18 at 16:52
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,562917
6,562917
add a comment |
add a comment |
$begingroup$
If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
$endgroup$
add a comment |
$begingroup$
If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
$endgroup$
add a comment |
$begingroup$
If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
$endgroup$
If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
answered Dec 5 '18 at 16:48
mlerma54mlerma54
1,177148
1,177148
add a comment |
add a comment |
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