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Proving that a sequence diverges to -inf
Clash Royale CLAN TAG#URR8PPP
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Prove that the divergence of the following sequence.
$$s_n= -2n^frac98$$
To prove this I was thinking of starting with $$n^frac98 > n$$
Then, $$2n^frac98 > 2n$$
Then, $$-2n^frac98 < -2n$$.
Then proving that $$lim_ntoinfty-2n = -infty$$
And since $$-2n^frac98 < -2n$$, by comparison, $$s_n= -2n^frac98$$ diverges.
I know there is more than one way to prove divergent of a sequence such as using contradiction of convergence using epsilon... Etc. But does my method count as a valid proof? I also tried proving this using the odd and even integer subsequence to prove this but I got stuck. Thanks.
real-analysis sequences-and-series
edited Jul 25 at 23:31
Holo
4,1072528
asked Jul 25 at 23:26
Donald Devy
63
add a comment |Â
up vote
1
down vote
favorite
Prove that the divergence of the following sequence.
$$s_n= -2n^frac98$$
To prove this I was thinking of starting with $$n^frac98 > n$$
Then, $$2n^frac98 > 2n$$
Then, $$-2n^frac98 < -2n$$.
Then proving that $$lim_ntoinfty-2n = -infty$$
And since $$-2n^frac98 < -2n$$, by comparison, $$s_n= -2n^frac98$$ diverges.
I know there is more than one way to prove divergent of a sequence such as using contradiction of convergence using epsilon... Etc. But does my method count as a valid proof? I also tried proving this using the odd and even integer subsequence to prove this but I got stuck. Thanks.
real-analysis sequences-and-series
edited Jul 25 at 23:31
Holo
4,1072528
asked Jul 25 at 23:26
Donald Devy
63
look here for mathjax tutorial
– Holo
Jul 25 at 23:28
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that the divergence of the following sequence.
$$s_n= -2n^frac98$$
To prove this I was thinking of starting with $$n^frac98 > n$$
Then, $$2n^frac98 > 2n$$
Then, $$-2n^frac98 < -2n$$.
Then proving that $$lim_ntoinfty-2n = -infty$$
And since $$-2n^frac98 < -2n$$, by comparison, $$s_n= -2n^frac98$$ diverges.
I know there is more than one way to prove divergent of a sequence such as using contradiction of convergence using epsilon... Etc. But does my method count as a valid proof? I also tried proving this using the odd and even integer subsequence to prove this but I got stuck. Thanks.
real-analysis sequences-and-series
edited Jul 25 at 23:31
Holo
4,1072528
asked Jul 25 at 23:26
Donald Devy
63
Prove that the divergence of the following sequence.
$$s_n= -2n^frac98$$
To prove this I was thinking of starting with $$n^frac98 > n$$
Then, $$2n^frac98 > 2n$$
Then, $$-2n^frac98 < -2n$$.
Then proving that $$lim_ntoinfty-2n = -infty$$
And since $$-2n^frac98 < -2n$$, by comparison, $$s_n= -2n^frac98$$ diverges.
I know there is more than one way to prove divergent of a sequence such as using contradiction of convergence using epsilon... Etc. But does my method count as a valid proof? I also tried proving this using the odd and even integer subsequence to prove this but I got stuck. Thanks.
real-analysis sequences-and-series
edited Jul 25 at 23:31
Holo
4,1072528
asked Jul 25 at 23:26
Donald Devy
63
edited Jul 25 at 23:31
Holo
4,1072528
edited Jul 25 at 23:31
Holo
4,1072528
edited Jul 25 at 23:31
Holo
4,1072528
4,1072528
asked Jul 25 at 23:26
Donald Devy
63
asked Jul 25 at 23:26
Donald Devy
63
asked Jul 25 at 23:26
Donald Devy
63
63
look here for mathjax tutorial
– Holo
Jul 25 at 23:28
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10
add a comment |Â
look here for mathjax tutorial
– Holo
Jul 25 at 23:28
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10
look here for mathjax tutorial
– Holo
Jul 25 at 23:28
look here for mathjax tutorial
– Holo
Jul 25 at 23:28
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n to -infty implies a_n to -infty$$ which you used implicitely in your proof.
answered Jul 26 at 7:12
Math_QED
6,35331344
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n to -infty implies a_n to -infty$$ which you used implicitely in your proof.
answered Jul 26 at 7:12
Math_QED
6,35331344
add a comment |Â
up vote
0
down vote
accepted
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n to -infty implies a_n to -infty$$ which you used implicitely in your proof.
answered Jul 26 at 7:12
Math_QED
6,35331344
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n to -infty implies a_n to -infty$$ which you used implicitely in your proof.
answered Jul 26 at 7:12
Math_QED
6,35331344
It is completely correct. In the comments, I provided a proof for the fact:
$$a_n < b_n, b_n to -infty implies a_n to -infty$$ which you used implicitely in your proof.
answered Jul 26 at 7:12
Math_QED
6,35331344
answered Jul 26 at 7:12
Math_QED
6,35331344
answered Jul 26 at 7:12
Math_QED
6,35331344
answered Jul 26 at 7:12
Math_QED
6,35331344
6,35331344
add a comment |Â
add a comment |Â
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look here for mathjax tutorial
– Holo
Jul 25 at 23:28
What you is completely valid but need a proof(if $a_n<b_n$ and $b_nto-infty$ then $a_nto-infty$ need a proof)
– Holo
Jul 25 at 23:34
No. $f(n) < g(n)$ and $g(n) to -infty$ does not imply $f(n) to -infty$ since $lim f(n)$ might not exist at all.
– xbh
Jul 26 at 1:22
It actually does. Let $M$ a real number. Choose $N$ s.t. $g(n) < M$ for all $n >N$. Then $f(n) < M$ for all $n > N$ and $f(n) to - infty$.
– Math_QED
Jul 26 at 7:10