Finding a sequence $a_n$ that diverges such that $|a_n|$ converges (in $mathbb{R}$)
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Finding a sequence $a_n$ that diverges such that $|a_n|$ converges (in $mathbb{R}$)
I am having a hard time finding an example that works. An example or hint would be greatly appreciated.
real-analysis sequences-and-series examples-counterexamples divergent-series absolute-convergence
edited Nov 29 at 3:16
Robert Howard
1,9181822
asked May 14 at 14:09
What Gives
274
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up vote
0
down vote
favorite
Finding a sequence $a_n$ that diverges such that $|a_n|$ converges (in $mathbb{R}$)
I am having a hard time finding an example that works. An example or hint would be greatly appreciated.
real-analysis sequences-and-series examples-counterexamples divergent-series absolute-convergence
edited Nov 29 at 3:16
Robert Howard
1,9181822
asked May 14 at 14:09
What Gives
274
1
What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Finding a sequence $a_n$ that diverges such that $|a_n|$ converges (in $mathbb{R}$)
I am having a hard time finding an example that works. An example or hint would be greatly appreciated.
real-analysis sequences-and-series examples-counterexamples divergent-series absolute-convergence
edited Nov 29 at 3:16
Robert Howard
1,9181822
asked May 14 at 14:09
What Gives
274
Finding a sequence $a_n$ that diverges such that $|a_n|$ converges (in $mathbb{R}$)
I am having a hard time finding an example that works. An example or hint would be greatly appreciated.
real-analysis sequences-and-series examples-counterexamples divergent-series absolute-convergence
real-analysis sequences-and-series examples-counterexamples divergent-series absolute-convergence
edited Nov 29 at 3:16
Robert Howard
1,9181822
asked May 14 at 14:09
What Gives
274
edited Nov 29 at 3:16
Robert Howard
1,9181822
asked May 14 at 14:09
What Gives
274
edited Nov 29 at 3:16
Robert Howard
1,9181822
edited Nov 29 at 3:16
Robert Howard
1,9181822
edited Nov 29 at 3:16
Robert Howard
1,9181822
1,9181822
asked May 14 at 14:09
What Gives
274
asked May 14 at 14:09
What Gives
274
asked May 14 at 14:09
What Gives
274
274
1
What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12
add a comment |
1
What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12
1
1
What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12
add a comment |
1 Answer
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accepted
Consider $a_n = (-1)^n$. It does not converge but $|a_n| = 1, forall ninmathbb{N}$.
answered May 14 at 14:13
mechanodroid
25.3k62245
add a comment |
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1 Answer
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accepted
Consider $a_n = (-1)^n$. It does not converge but $|a_n| = 1, forall ninmathbb{N}$.
answered May 14 at 14:13
mechanodroid
25.3k62245
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up vote
2
down vote
accepted
Consider $a_n = (-1)^n$. It does not converge but $|a_n| = 1, forall ninmathbb{N}$.
answered May 14 at 14:13
mechanodroid
25.3k62245
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Consider $a_n = (-1)^n$. It does not converge but $|a_n| = 1, forall ninmathbb{N}$.
answered May 14 at 14:13
mechanodroid
25.3k62245
Consider $a_n = (-1)^n$. It does not converge but $|a_n| = 1, forall ninmathbb{N}$.
answered May 14 at 14:13
mechanodroid
25.3k62245
answered May 14 at 14:13
mechanodroid
25.3k62245
answered May 14 at 14:13
mechanodroid
25.3k62245
answered May 14 at 14:13
mechanodroid
25.3k62245
25.3k62245
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add a comment |
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What norm is $||.||$ - do you just mean the absolute value? If so, just consider $a_n = frac{1}{n}$
– PhysicsMathsLove
May 14 at 14:11
A hint is that I think they mean "the limit of $a_n$ does not exist" or "$a_n$ does not converge" when they say "$a_n$ diverges." (I typically think of something that "diverges" as something that goes to infinity in some sense, but that is not happening here)
– Michael
May 14 at 14:12