Finding a Divergent series for a convergent series.
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I was asked that, giving $$sum_{n=1}^infty a_n,$$ find $a_n$ such that the series diverges, yet
$$a_nlefrac{1}{2^n}$$
I came up with this: $$sum_{i=1}^n-frac{1}{n}$$ My logic is that I know $1/n$ would diverge, and I know it would be less than $1/2^n$ if I put a negative sign. Am I correct? If not, what value for an would solve the problem?
sequences-and-series convergence
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
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add a comment |
$begingroup$
I was asked that, giving $$sum_{n=1}^infty a_n,$$ find $a_n$ such that the series diverges, yet
$$a_nlefrac{1}{2^n}$$
I came up with this: $$sum_{i=1}^n-frac{1}{n}$$ My logic is that I know $1/n$ would diverge, and I know it would be less than $1/2^n$ if I put a negative sign. Am I correct? If not, what value for an would solve the problem?
sequences-and-series convergence
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
$endgroup$
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Looks good to me!
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– SmileyCraft
Dec 16 '18 at 18:51
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If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53
add a comment |
$begingroup$
I was asked that, giving $$sum_{n=1}^infty a_n,$$ find $a_n$ such that the series diverges, yet
$$a_nlefrac{1}{2^n}$$
I came up with this: $$sum_{i=1}^n-frac{1}{n}$$ My logic is that I know $1/n$ would diverge, and I know it would be less than $1/2^n$ if I put a negative sign. Am I correct? If not, what value for an would solve the problem?
sequences-and-series convergence
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
$endgroup$
I was asked that, giving $$sum_{n=1}^infty a_n,$$ find $a_n$ such that the series diverges, yet
$$a_nlefrac{1}{2^n}$$
I came up with this: $$sum_{i=1}^n-frac{1}{n}$$ My logic is that I know $1/n$ would diverge, and I know it would be less than $1/2^n$ if I put a negative sign. Am I correct? If not, what value for an would solve the problem?
sequences-and-series convergence
sequences-and-series convergence
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
edited Dec 16 '18 at 18:51
Will Fisher
4,0331032
4,0331032
asked Dec 16 '18 at 18:47
A.B.A.B.
61
asked Dec 16 '18 at 18:47
A.B.A.B.
61
asked Dec 16 '18 at 18:47
A.B.A.B.
61
61
$begingroup$
Looks good to me!
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:51
$begingroup$
If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53
add a comment |
$begingroup$
Looks good to me!
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:51
$begingroup$
If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53
$begingroup$
Looks good to me!
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:51
$begingroup$
Looks good to me!
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:51
$begingroup$
If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53
$begingroup$
If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are correct. If we had that
$$|a_n|le frac{1}{2^n}$$
then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are correct. If we had that
$$|a_n|le frac{1}{2^n}$$
then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
$endgroup$
add a comment |
$begingroup$
You are correct. If we had that
$$|a_n|le frac{1}{2^n}$$
then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
$endgroup$
add a comment |
$begingroup$
You are correct. If we had that
$$|a_n|le frac{1}{2^n}$$
then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
$endgroup$
You are correct. If we had that
$$|a_n|le frac{1}{2^n}$$
then the series would necessarily converge by the comparison test. Thus the only way around this is to have $|a_n|$ large but $a_n$ negative.
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
answered Dec 16 '18 at 18:52
Will FisherWill Fisher
4,0331032
4,0331032
add a comment |
add a comment |
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$begingroup$
Looks good to me!
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:51
$begingroup$
If $a_n$ can be negative, then why not simply go with $a_n=-1$ ?
$endgroup$
– Sauhard Sharma
Dec 16 '18 at 18:53