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Real Analysis: Provide example of two Series that…

Don't understand inequality in order to prove Algebraic Limit TheoremIf two real series $sum a_n$ and $sum b_n$ converge but $sum a_n b_n$ does not can $sum a_n$ or $sum b_n$ converge absolutely?real analysis: $sum a_n$ converges if and only if $sum n a_{n^2}$ convergesWhy can't I apply the Dirichlet test here?IF $sum a_n$ and $sum b_n$ converge does $sum a_n + b_n$ convergeHelp with Real Analysis proofLet ${a_n}$ be a sequence of reals going to $0$ but whose series diverges. Prove that we can flip the signs of each term to get a convergent seriesFor which values of $x$ does the following series converge?Proof explanation: Convergence of a seriesSeries divergence

3

$begingroup$

I've posted the solution for this problem and I'm trying to understand this.

In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)

And why do we have to add one to the sum of partial sums?

QUESTION AND SOLUTION:

real-analysis

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asked 2 hours ago

K KAK KA

454

$endgroup$

add a comment | 

3

$begingroup$

I've posted the solution for this problem and I'm trying to understand this.

In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)

And why do we have to add one to the sum of partial sums?

QUESTION AND SOLUTION:

real-analysis

share|cite|improve this question

asked 2 hours ago

K KAK KA

454

$endgroup$

add a comment | 

$begingroup$

I've posted the solution for this problem and I'm trying to understand this.

In the end of the solution provided it says to continue this process. So, do we hold $a_n$ to be $frac{1}{n^2}$ and $b_n$ to be $frac{1}{900^2}$ for the next $900^2$ terms? And then hold $b_n$ to be $frac{1}{n^2}$ and $a_n$ to be $frac{1}{810000 ^2}$ for the next $810000 ^2$ terms? (because $900^2$ is $810000$)

And why do we have to add one to the sum of partial sums?

QUESTION AND SOLUTION:

real-analysis

share|cite|improve this question

asked 2 hours ago

K KAK KA

454

$endgroup$

share|cite|improve this question

asked 2 hours ago

K KAK KA

454

share|cite|improve this question

asked 2 hours ago

K KAK KA

454

asked 2 hours ago

K KAK KA

454

asked 2 hours ago

K KAK KA

454

2 Answers
2

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oldest

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3

$begingroup$

Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series
begin{matrix}
a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
end{matrix}
and modify them like so:
begin{matrix}
a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
end{matrix}
The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.

share|cite|improve this answer

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

$endgroup$

add a comment | 

3

$begingroup$

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.

To find how many terms we add again, think about the pattern
begin{align*}
(1+1) - 1 &= 1^2 \
(5 + 1) - 2 &= 2^2 \
(30 + 1) - 6 &= 5^2 \
(930 + 1) - 31 &= 30^2 \
(865830 + 1) - 931 &= 930^2 \
(x + 1) - 865831 &= 865830^2 \
dotsb
end{align*}

share|cite|improve this answer

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

$endgroup$

add a comment | 

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3

$begingroup$

Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series
begin{matrix}
a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
end{matrix}
and modify them like so:
begin{matrix}
a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
end{matrix}
The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.

share|cite|improve this answer

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

$endgroup$

add a comment | 

3

$begingroup$

Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series
begin{matrix}
a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
end{matrix}
and modify them like so:
begin{matrix}
a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
end{matrix}
The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.

share|cite|improve this answer

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

$endgroup$

add a comment | 

$begingroup$

Essentially, they're making $a_n$ and $b_n$ a positive monotone summable sequence (in this case, $frac{1}{n^2}$), and just "pausing" each sequence long enough that a $1$ is added to the partial sum, thereby forcing the sum to diverge.

So, start with the same series
begin{matrix}
a_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & frac{1}{5^2} & frac{1}{6^2} & frac{1}{7^2} & frac{1}{8^2} & frac{1}{9^2} & frac{1}{10^2} & frac{1}{11^2} & frac{1}{12^2} & frac{1}{13^2} & frac{1}{14^2} & frac{1}{15^2} & frac{1}{16^2} & cdots &bigr)
end{matrix}
and modify them like so:
begin{matrix}
a_n = bigl(1 & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & color{red}{frac{1}{4}} & frac{1}{6^2} & frac{1}{7^2} & cdots & frac{1}{29^2} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & cdots & color{red}{frac{1}{900}} & color{red}{frac{1}{900}} & frac{1}{930^2} & cdots &bigr) \
b_n = bigl(1 & frac{1}{2^2} & frac{1}{3^2} & frac{1}{4^2} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & cdots & color{red}{frac{1}{25}} & color{red}{frac{1}{25}} & frac{1}{31^2} & frac{1}{32^2} & cdots & frac{1}{928^2} & color{red}{frac{1}{929^2}} & color{red}{frac{1}{929^2}} & cdots &bigr).
end{matrix}
The streaks of red numbers add to $1$, and occur infinitely many often in both sequences, so each partial sum becomes unbounded and hence the series fails to converge. But, the minimum of the two sequences is always $frac{1}{n^2}$.

share|cite|improve this answer

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

answered 1 hour ago

Theo BenditTheo Bendit

18.8k12253

3

$begingroup$

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.

To find how many terms we add again, think about the pattern
begin{align*}
(1+1) - 1 &= 1^2 \
(5 + 1) - 2 &= 2^2 \
(30 + 1) - 6 &= 5^2 \
(930 + 1) - 31 &= 30^2 \
(865830 + 1) - 931 &= 930^2 \
(x + 1) - 865831 &= 865830^2 \
dotsb
end{align*}

share|cite|improve this answer

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

$endgroup$

add a comment | 

3

$begingroup$

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.

To find how many terms we add again, think about the pattern
begin{align*}
(1+1) - 1 &= 1^2 \
(5 + 1) - 2 &= 2^2 \
(30 + 1) - 6 &= 5^2 \
(930 + 1) - 31 &= 30^2 \
(865830 + 1) - 931 &= 930^2 \
(x + 1) - 865831 &= 865830^2 \
dotsb
end{align*}

share|cite|improve this answer

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

$endgroup$

add a comment | 

$begingroup$

The idea behind the more challenging version is to construct $(a_n)$ and $(b_n)$ such that $sum a_n,sum b_n$ both diverge, but $sum min{a_n,b_n}$ converges. The way the author of this solution has chosen to proceed is by making $(a_n)$ and $(b_n)$ such that for each $n$, we have $min{a_n,b_n} = 1/n^2$, and yet we add enough small constant terms to each sequence so that the partial sums eventually grow by $1$ if we wait long enough. This growth by $1$ repeated over and over again ensures that the series $sum a_n,sum b_n$ both diverge since their partial sums each grow without bound by merely waiting long enough.

To find how many terms we add again, think about the pattern
begin{align*}
(1+1) - 1 &= 1^2 \
(5 + 1) - 2 &= 2^2 \
(30 + 1) - 6 &= 5^2 \
(930 + 1) - 31 &= 30^2 \
(865830 + 1) - 931 &= 930^2 \
(x + 1) - 865831 &= 865830^2 \
dotsb
end{align*}

share|cite|improve this answer

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441

answered 1 hour ago

Alex OrtizAlex Ortiz

10.8k21441