Mixing
Prove that the limit of a sequence is 0. (Stolz-Cesaro theorem?) [duplicate]
This question already has an answer here:
$(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ [duplicate]
2 answers
Prove that if $(a_{n+1} + a_n) to 0$, then $frac{a_n}{n} to 0$. Is it possible to replace $0$ by some $g$?
It looks like using Stolz-Cesaro theorem is required here but I don't really have any ideas on how to start.
limits analysis
edited Nov 20 '18 at 23:58
gimusi
1
asked Nov 20 '18 at 23:24
KacperR
295
marked as duplicate by Martin R, Brahadeesh, amWhy, John B, jgon Nov 21 '18 at 16:31
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This question already has an answer here:
$(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ [duplicate]
2 answers
Prove that if $(a_{n+1} + a_n) to 0$, then $frac{a_n}{n} to 0$. Is it possible to replace $0$ by some $g$?
It looks like using Stolz-Cesaro theorem is required here but I don't really have any ideas on how to start.
limits analysis
edited Nov 20 '18 at 23:58
gimusi
1
asked Nov 20 '18 at 23:24
KacperR
295
marked as duplicate by Martin R, Brahadeesh, amWhy, John B, jgon Nov 21 '18 at 16:31
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.
add a comment |
This question already has an answer here:
$(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ [duplicate]
2 answers
Prove that if $(a_{n+1} + a_n) to 0$, then $frac{a_n}{n} to 0$. Is it possible to replace $0$ by some $g$?
It looks like using Stolz-Cesaro theorem is required here but I don't really have any ideas on how to start.
limits analysis
edited Nov 20 '18 at 23:58
gimusi
1
asked Nov 20 '18 at 23:24
KacperR
295
This question already has an answer here:
$(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ [duplicate]
2 answers
Prove that if $(a_{n+1} + a_n) to 0$, then $frac{a_n}{n} to 0$. Is it possible to replace $0$ by some $g$?
It looks like using Stolz-Cesaro theorem is required here but I don't really have any ideas on how to start.
This question already has an answer here:
$(a_n+a_{n+1})$ is convergent implies $(a_n/n)$ converges to $0$ [duplicate]
2 answers
limits analysis
limits analysis
edited Nov 20 '18 at 23:58
gimusi
1
asked Nov 20 '18 at 23:24
KacperR
295
edited Nov 20 '18 at 23:58
gimusi
1
asked Nov 20 '18 at 23:24
KacperR
295
edited Nov 20 '18 at 23:58
gimusi
1
edited Nov 20 '18 at 23:58
gimusi
1
edited Nov 20 '18 at 23:58
gimusi
1
1
asked Nov 20 '18 at 23:24
KacperR
295
asked Nov 20 '18 at 23:24
KacperR
295
asked Nov 20 '18 at 23:24
KacperR
295
295
marked as duplicate by Martin R, Brahadeesh, amWhy, John B, jgon Nov 21 '18 at 16:31
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 Martin R, Brahadeesh, amWhy, John B, jgon Nov 21 '18 at 16:31
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.
add a comment |
add a comment |
1 Answer
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Let consider
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}$$
then by Stolz-Cesaro
$$frac{b_{n+1}-b_n}{c_{n+1}-c_n}=(-1)^{n+1}a_{n+1}-(-1)^na_n=-(-1)^n(a_{n+1}+a_n)to 0$$
and therefore
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}to 0 implies frac{a_n}{n}to 0$$
answered Nov 20 '18 at 23:29
gimusi
1
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let consider
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}$$
then by Stolz-Cesaro
$$frac{b_{n+1}-b_n}{c_{n+1}-c_n}=(-1)^{n+1}a_{n+1}-(-1)^na_n=-(-1)^n(a_{n+1}+a_n)to 0$$
and therefore
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}to 0 implies frac{a_n}{n}to 0$$
answered Nov 20 '18 at 23:29
gimusi
1
add a comment |
Let consider
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}$$
then by Stolz-Cesaro
$$frac{b_{n+1}-b_n}{c_{n+1}-c_n}=(-1)^{n+1}a_{n+1}-(-1)^na_n=-(-1)^n(a_{n+1}+a_n)to 0$$
and therefore
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}to 0 implies frac{a_n}{n}to 0$$
answered Nov 20 '18 at 23:29
gimusi
1
add a comment |
Let consider
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}$$
then by Stolz-Cesaro
$$frac{b_{n+1}-b_n}{c_{n+1}-c_n}=(-1)^{n+1}a_{n+1}-(-1)^na_n=-(-1)^n(a_{n+1}+a_n)to 0$$
and therefore
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}to 0 implies frac{a_n}{n}to 0$$
answered Nov 20 '18 at 23:29
gimusi
1
Let consider
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}$$
then by Stolz-Cesaro
$$frac{b_{n+1}-b_n}{c_{n+1}-c_n}=(-1)^{n+1}a_{n+1}-(-1)^na_n=-(-1)^n(a_{n+1}+a_n)to 0$$
and therefore
$$frac{b_n}{c_n}=frac{(-1)^na_n}{n}to 0 implies frac{a_n}{n}to 0$$
answered Nov 20 '18 at 23:29
gimusi
1
answered Nov 20 '18 at 23:29
gimusi
1
answered Nov 20 '18 at 23:29
gimusi
1
answered Nov 20 '18 at 23:29
gimusi
1
1
add a comment |
add a comment |
