Prove/Disprove $sum a_n$ and $sum b_n$ converge together iff $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n...
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True/False: $mathop {lim }limits_{n to infty } {{{a_n}} over {{b_n}}} = 1$ implies $sum {{a_n},sum {{b_n}} } $ converge or diverge together.
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Prove/Disprove that if $sum a_n$ and $sum b_n$ are some series and $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ then the series converge together or not converge together.
This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?)
calculus sequences-and-series limits convergence
asked Dec 5 at 12:49
Boaz Yakubov
254
marked as duplicate by Chinnapparaj R, Cm7F7Bb, Community♦ Dec 5 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
True/False: $mathop {lim }limits_{n to infty } {{{a_n}} over {{b_n}}} = 1$ implies $sum {{a_n},sum {{b_n}} } $ converge or diverge together.
3 answers
Prove/Disprove that if $sum a_n$ and $sum b_n$ are some series and $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ then the series converge together or not converge together.
This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?)
calculus sequences-and-series limits convergence
asked Dec 5 at 12:49
Boaz Yakubov
254
marked as duplicate by Chinnapparaj R, Cm7F7Bb, Community♦ Dec 5 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11
add a comment |
up vote
0
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favorite
up vote
0
down vote
favorite
This question already has an answer here:
True/False: $mathop {lim }limits_{n to infty } {{{a_n}} over {{b_n}}} = 1$ implies $sum {{a_n},sum {{b_n}} } $ converge or diverge together.
3 answers
Prove/Disprove that if $sum a_n$ and $sum b_n$ are some series and $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ then the series converge together or not converge together.
This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?)
calculus sequences-and-series limits convergence
asked Dec 5 at 12:49
Boaz Yakubov
254
This question already has an answer here:
True/False: $mathop {lim }limits_{n to infty } {{{a_n}} over {{b_n}}} = 1$ implies $sum {{a_n},sum {{b_n}} } $ converge or diverge together.
3 answers
Prove/Disprove that if $sum a_n$ and $sum b_n$ are some series and $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ then the series converge together or not converge together.
This doesn't seem to be correct to me so maybe there is some counter example. i know this is true if both series are strictly nonnegative (from the first series comparison test) - how does it change if both are negative (or alternate?)
This question already has an answer here:
True/False: $mathop {lim }limits_{n to infty } {{{a_n}} over {{b_n}}} = 1$ implies $sum {{a_n},sum {{b_n}} } $ converge or diverge together.
3 answers
calculus sequences-and-series limits convergence
calculus sequences-and-series limits convergence
asked Dec 5 at 12:49
Boaz Yakubov
254
asked Dec 5 at 12:49
Boaz Yakubov
254
asked Dec 5 at 12:49
Boaz Yakubov
254
asked Dec 5 at 12:49
Boaz Yakubov
254
asked Dec 5 at 12:49
Boaz Yakubov
254
254
marked as duplicate by Chinnapparaj R, Cm7F7Bb, Community♦ Dec 5 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Chinnapparaj R, Cm7F7Bb, Community♦ Dec 5 at 13:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11
add a comment |
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11
add a comment |
2 Answers
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If $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ ,
then $$(exists Nin Bbb N, forall n > N):; frac{1}{2}a_n le b_nle frac{3}{2}a_n$$
Thus if $sum a_n$ converges then so does $sum b_n$ and vice versa
Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below
answered Dec 5 at 12:52
TheD0ubleT
37718
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
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0
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A counterexample is:
$$a_n = frac{(-1)^n}{sqrt{n}} text{ and } b_n = frac{(-1)^n}{sqrt{n} + (-1)^n}.$$
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
If $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ ,
then $$(exists Nin Bbb N, forall n > N):; frac{1}{2}a_n le b_nle frac{3}{2}a_n$$
Thus if $sum a_n$ converges then so does $sum b_n$ and vice versa
Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below
answered Dec 5 at 12:52
TheD0ubleT
37718
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
add a comment |
up vote
4
down vote
If $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ ,
then $$(exists Nin Bbb N, forall n > N):; frac{1}{2}a_n le b_nle frac{3}{2}a_n$$
Thus if $sum a_n$ converges then so does $sum b_n$ and vice versa
Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below
answered Dec 5 at 12:52
TheD0ubleT
37718
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
add a comment |
up vote
4
down vote
up vote
4
down vote
If $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ ,
then $$(exists Nin Bbb N, forall n > N):; frac{1}{2}a_n le b_nle frac{3}{2}a_n$$
Thus if $sum a_n$ converges then so does $sum b_n$ and vice versa
Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below
answered Dec 5 at 12:52
TheD0ubleT
37718
If $underset{nto infty }{mathop{lim }},{{a}_{n}/b_n } = 1$ ,
then $$(exists Nin Bbb N, forall n > N):; frac{1}{2}a_n le b_nle frac{3}{2}a_n$$
Thus if $sum a_n$ converges then so does $sum b_n$ and vice versa
Update As stated in the comment this is infact only true if $a_n$ and $b_n$ keep a constant sign (positive or negative) after a certain n. Otherwise it's false: cf. the counter example given below
answered Dec 5 at 12:52
TheD0ubleT
37718
edited Dec 5 at 16:11
answered Dec 5 at 12:52
TheD0ubleT
37718
answered Dec 5 at 12:52
TheD0ubleT
37718
answered Dec 5 at 12:52
TheD0ubleT
37718
37718
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
add a comment |
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
1
1
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
@A.Pongrácz For series of positive numbers this is a complete proof. In the example you give, both series diverge.
– MaoWao
Dec 5 at 13:09
add a comment |
up vote
0
down vote
A counterexample is:
$$a_n = frac{(-1)^n}{sqrt{n}} text{ and } b_n = frac{(-1)^n}{sqrt{n} + (-1)^n}.$$
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
add a comment |
up vote
0
down vote
A counterexample is:
$$a_n = frac{(-1)^n}{sqrt{n}} text{ and } b_n = frac{(-1)^n}{sqrt{n} + (-1)^n}.$$
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
add a comment |
up vote
0
down vote
up vote
0
down vote
A counterexample is:
$$a_n = frac{(-1)^n}{sqrt{n}} text{ and } b_n = frac{(-1)^n}{sqrt{n} + (-1)^n}.$$
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
A counterexample is:
$$a_n = frac{(-1)^n}{sqrt{n}} text{ and } b_n = frac{(-1)^n}{sqrt{n} + (-1)^n}.$$
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
edited Dec 5 at 13:34
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
answered Dec 5 at 12:59
mathcounterexamples.net
23.9k21753
23.9k21753
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
add a comment |
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
You're right I fixed it.
– mathcounterexamples.net
Dec 5 at 13:37
add a comment |
This is true for series with a posiitve (or negative) general term. It is the criterion by equivalence of functions.
– Bernard
Dec 5 at 13:00
Thank you very much everyone for your time, turns out it was a duplicate..
– Boaz Yakubov
Dec 5 at 13:11