Mixing
Upper Bound of a Subseries of a Positive Convergent Sequence
$begingroup$
Let $(a_n) to 0$ and that $a_n >0, forall n$. I am trying to show that:
$$
exists (a_{n_k}): sum_{k=1}^{infty}a_{n_k}<1
$$
Thought: I can show that $(a_n)$ has a strictly decreasing subsequence. Because $(a_n) to 0$, this subsequence can be constructed to have an upper bound as small as needed. I think this has to do with this question. Other than that, I am not sure how to tackle the problem.
real-analysis sequences-and-series
edited Jan 17 at 19:25
Foobaz John
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
$endgroup$
add a comment |
$begingroup$
Let $(a_n) to 0$ and that $a_n >0, forall n$. I am trying to show that:
$$
exists (a_{n_k}): sum_{k=1}^{infty}a_{n_k}<1
$$
Thought: I can show that $(a_n)$ has a strictly decreasing subsequence. Because $(a_n) to 0$, this subsequence can be constructed to have an upper bound as small as needed. I think this has to do with this question. Other than that, I am not sure how to tackle the problem.
real-analysis sequences-and-series
edited Jan 17 at 19:25
Foobaz John
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
$endgroup$
add a comment |
$begingroup$
Let $(a_n) to 0$ and that $a_n >0, forall n$. I am trying to show that:
$$
exists (a_{n_k}): sum_{k=1}^{infty}a_{n_k}<1
$$
Thought: I can show that $(a_n)$ has a strictly decreasing subsequence. Because $(a_n) to 0$, this subsequence can be constructed to have an upper bound as small as needed. I think this has to do with this question. Other than that, I am not sure how to tackle the problem.
real-analysis sequences-and-series
edited Jan 17 at 19:25
Foobaz John
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
$endgroup$
Let $(a_n) to 0$ and that $a_n >0, forall n$. I am trying to show that:
$$
exists (a_{n_k}): sum_{k=1}^{infty}a_{n_k}<1
$$
Thought: I can show that $(a_n)$ has a strictly decreasing subsequence. Because $(a_n) to 0$, this subsequence can be constructed to have an upper bound as small as needed. I think this has to do with this question. Other than that, I am not sure how to tackle the problem.
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 17 at 19:25
Foobaz John
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
edited Jan 17 at 19:25
Foobaz John
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
edited Jan 17 at 19:25
Foobaz John
22.2k41452
edited Jan 17 at 19:25
Foobaz John
22.2k41452
edited Jan 17 at 19:25
Foobaz John
22.2k41452
22.2k41452
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
asked Jan 17 at 18:56
A Slow LearnerA Slow Learner
453212
453212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
Since $a_nto 0$, there exists $n_1<n_2<dotsb$ such that
$$
0<a_{n_k}< frac{1}{4^k}.
$$
Now sum over $k$ and use the formula for a geometric series.
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Since $a_nto 0$, there exists $n_1<n_2<dotsb$ such that
$$
0<a_{n_k}< frac{1}{4^k}.
$$
Now sum over $k$ and use the formula for a geometric series.
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
add a comment |
$begingroup$
Hint
Since $a_nto 0$, there exists $n_1<n_2<dotsb$ such that
$$
0<a_{n_k}< frac{1}{4^k}.
$$
Now sum over $k$ and use the formula for a geometric series.
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
add a comment |
$begingroup$
Hint
Since $a_nto 0$, there exists $n_1<n_2<dotsb$ such that
$$
0<a_{n_k}< frac{1}{4^k}.
$$
Now sum over $k$ and use the formula for a geometric series.
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
Hint
Since $a_nto 0$, there exists $n_1<n_2<dotsb$ such that
$$
0<a_{n_k}< frac{1}{4^k}.
$$
Now sum over $k$ and use the formula for a geometric series.
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
answered Jan 17 at 19:24
Foobaz JohnFoobaz John
22.2k41452
22.2k41452
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
add a comment |
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
$begingroup$
Thank you! Just a related question though: here you use $frac{1}{4^k}$, but we can underestimate the sum by using an even smaller number, correct? Similarly, if we use $frac{1}{2^k}$, then we can obtain a tighter bound (=1). Also, is there a general result behind this particular question?
$endgroup$
– A Slow Learner
Jan 17 at 21:06
add a comment |
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