How this series a_j converges? [closed]
We have $a_ngeq 0$ and suppose that $sum_{j=n}^{2n} a_jleq frac{1}{sqrt{n}}$. I dont know how to derive that $sum a_j$ converges.
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 23:32
Tsubaki
14
closed as off-topic by Saad, RRL, KReiser, DRF, Gibbs Dec 10 '18 at 10:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, KReiser, DRF, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
We have $a_ngeq 0$ and suppose that $sum_{j=n}^{2n} a_jleq frac{1}{sqrt{n}}$. I dont know how to derive that $sum a_j$ converges.
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 23:32
Tsubaki
14
closed as off-topic by Saad, RRL, KReiser, DRF, Gibbs Dec 10 '18 at 10:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, KReiser, DRF, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00
add a comment |
We have $a_ngeq 0$ and suppose that $sum_{j=n}^{2n} a_jleq frac{1}{sqrt{n}}$. I dont know how to derive that $sum a_j$ converges.
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 23:32
Tsubaki
14
We have $a_ngeq 0$ and suppose that $sum_{j=n}^{2n} a_jleq frac{1}{sqrt{n}}$. I dont know how to derive that $sum a_j$ converges.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 23:32
Tsubaki
14
asked Dec 9 '18 at 23:32
Tsubaki
14
asked Dec 9 '18 at 23:32
Tsubaki
14
asked Dec 9 '18 at 23:32
Tsubaki
14
asked Dec 9 '18 at 23:32
Tsubaki
14
14
closed as off-topic by Saad, RRL, KReiser, DRF, Gibbs Dec 10 '18 at 10:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, KReiser, DRF, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, RRL, KReiser, DRF, Gibbs Dec 10 '18 at 10:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, KReiser, DRF, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00
add a comment |
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00
add a comment |
2 Answers
2
active
oldest
votes
We have that
$$sum_{j=1}^{infty} a_j=sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}-1} a_jright)le sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}} a_jright)le sum_{k=0}^{infty}frac{1}{sqrt{2^k}}$$
answered Dec 9 '18 at 23:48
gimusi
1
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
|
show 10 more comments
Let $$b_n = sum_{j = 2^{n-1}}^{2^n-1} a_j leqfrac{1}{sqrt{2^n}}$$ Then $sum b_n = sum a_n$ and $sum b_n$ converges as $sum frac{1}{sqrt{2^n}}$ converges.
answered Dec 9 '18 at 23:38
ODF
1,326410
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We have that
$$sum_{j=1}^{infty} a_j=sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}-1} a_jright)le sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}} a_jright)le sum_{k=0}^{infty}frac{1}{sqrt{2^k}}$$
answered Dec 9 '18 at 23:48
gimusi
1
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
|
show 10 more comments
We have that
$$sum_{j=1}^{infty} a_j=sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}-1} a_jright)le sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}} a_jright)le sum_{k=0}^{infty}frac{1}{sqrt{2^k}}$$
answered Dec 9 '18 at 23:48
gimusi
1
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
|
show 10 more comments
We have that
$$sum_{j=1}^{infty} a_j=sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}-1} a_jright)le sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}} a_jright)le sum_{k=0}^{infty}frac{1}{sqrt{2^k}}$$
answered Dec 9 '18 at 23:48
gimusi
1
We have that
$$sum_{j=1}^{infty} a_j=sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}-1} a_jright)le sum_{k=0}^{infty}left(sum_{j=2^k}^{2^{k+1}} a_jright)le sum_{k=0}^{infty}frac{1}{sqrt{2^k}}$$
answered Dec 9 '18 at 23:48
gimusi
1
edited Dec 9 '18 at 23:54
answered Dec 9 '18 at 23:48
gimusi
1
answered Dec 9 '18 at 23:48
gimusi
1
answered Dec 9 '18 at 23:48
gimusi
1
1
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
|
show 10 more comments
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
How does the first equality derive?
– Tsubaki
Dec 10 '18 at 0:00
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
@Tsubaki Let try with $k=0,1,2...$ and see what happens.
– gimusi
Dec 10 '18 at 0:01
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality?
– Tsubaki
Dec 10 '18 at 0:11
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
@Tsubaki for $k=0 implies j=1$ for $k=1 implies j=2 to 3$ for $k=2 implies j=4 to 7$ and so on.Can you see the pattern?
– gimusi
Dec 10 '18 at 0:16
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
|
show 10 more comments
Let $$b_n = sum_{j = 2^{n-1}}^{2^n-1} a_j leqfrac{1}{sqrt{2^n}}$$ Then $sum b_n = sum a_n$ and $sum b_n$ converges as $sum frac{1}{sqrt{2^n}}$ converges.
answered Dec 9 '18 at 23:38
ODF
1,326410
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
add a comment |
Let $$b_n = sum_{j = 2^{n-1}}^{2^n-1} a_j leqfrac{1}{sqrt{2^n}}$$ Then $sum b_n = sum a_n$ and $sum b_n$ converges as $sum frac{1}{sqrt{2^n}}$ converges.
answered Dec 9 '18 at 23:38
ODF
1,326410
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
add a comment |
Let $$b_n = sum_{j = 2^{n-1}}^{2^n-1} a_j leqfrac{1}{sqrt{2^n}}$$ Then $sum b_n = sum a_n$ and $sum b_n$ converges as $sum frac{1}{sqrt{2^n}}$ converges.
answered Dec 9 '18 at 23:38
ODF
1,326410
Let $$b_n = sum_{j = 2^{n-1}}^{2^n-1} a_j leqfrac{1}{sqrt{2^n}}$$ Then $sum b_n = sum a_n$ and $sum b_n$ converges as $sum frac{1}{sqrt{2^n}}$ converges.
answered Dec 9 '18 at 23:38
ODF
1,326410
edited Dec 9 '18 at 23:50
answered Dec 9 '18 at 23:38
ODF
1,326410
answered Dec 9 '18 at 23:38
ODF
1,326410
answered Dec 9 '18 at 23:38
ODF
1,326410
1,326410
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
add a comment |
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
May I ask how latter part converges?
– Tsubaki
Dec 10 '18 at 0:55
add a comment |
Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge.
– GenericMathematician
Dec 9 '18 at 23:35
The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy.
– Sean Roberson
Dec 10 '18 at 0:00