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Finding a converging series, which when multiplied by another, diverges
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Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Sequences and series and the alternating series testHow can I show whether the series $sumlimits_{n=1}^infty frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges?Application of Stirling's theorem for the given seriesConvergence of $ sum_{n=1}^inftyleft(exp(frac{(-1)^n}{n})-1right)$?Sequence $(a_n)$ s.t. $sum a_n$ converges and $|{k in {0,dots, n}: a_{k+1} > a_k}|/n to 1$?If we think of infinity as a number, how does it affect the compactness/completeness of a metric space?What is the reasoning behind why the ratio test works?Does $ sum^{infty}_{3} frac{n^2}{(ln(ln(n)))^{ln(n)}} $ converge?Finding a sequence $a_n$ such that $sum a_n$ converges but $sum a_n^3$ diverges.Tools for finding bounds on power seriesSequences and series and the alternating series test
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I was thinking of using some sort of an alternating sequence for $(a_{n})_{n=1}$, but wasn't sure where to go. I know that an alternating series is not divergent, it's just not convergent since it doesn't tend to infinity but rather oscillates between two numbers.
Any ideas are appreciated.
Link to previously asked question: Sequences and series and the alternating series test
If the solutions are correct, can someone explain them properly please.
real-analysis analysis
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add a comment |
$begingroup$
I was thinking of using some sort of an alternating sequence for $(a_{n})_{n=1}$, but wasn't sure where to go. I know that an alternating series is not divergent, it's just not convergent since it doesn't tend to infinity but rather oscillates between two numbers.
Any ideas are appreciated.
Link to previously asked question: Sequences and series and the alternating series test
If the solutions are correct, can someone explain them properly please.
real-analysis analysis
$endgroup$
$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
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– Oscar Lanzi
Mar 25 at 10:07
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@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29
add a comment |
$begingroup$
I was thinking of using some sort of an alternating sequence for $(a_{n})_{n=1}$, but wasn't sure where to go. I know that an alternating series is not divergent, it's just not convergent since it doesn't tend to infinity but rather oscillates between two numbers.
Any ideas are appreciated.
Link to previously asked question: Sequences and series and the alternating series test
If the solutions are correct, can someone explain them properly please.
real-analysis analysis
$endgroup$
I was thinking of using some sort of an alternating sequence for $(a_{n})_{n=1}$, but wasn't sure where to go. I know that an alternating series is not divergent, it's just not convergent since it doesn't tend to infinity but rather oscillates between two numbers.
Any ideas are appreciated.
Link to previously asked question: Sequences and series and the alternating series test
If the solutions are correct, can someone explain them properly please.
real-analysis analysis
real-analysis analysis
edited Apr 15 at 10:40
asked Mar 25 at 9:59
user437703
$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
$endgroup$
– Oscar Lanzi
Mar 25 at 10:07
$begingroup$
@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29
add a comment |
$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
$endgroup$
– Oscar Lanzi
Mar 25 at 10:07
$begingroup$
@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29
$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
$endgroup$
– Oscar Lanzi
Mar 25 at 10:07
$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
$endgroup$
– Oscar Lanzi
Mar 25 at 10:07
$begingroup$
@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
We can find integers $n_1<n_2<cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n notin {n_1,n_2,cdots}$ and $a_{n_k}=q_{n_k}^{-1}$. $sum a_n =sum a_{n_k} <sum frac 1 {2^{k}} <infty$ and $sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
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add a comment |
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$begingroup$
We can find integers $n_1<n_2<cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n notin {n_1,n_2,cdots}$ and $a_{n_k}=q_{n_k}^{-1}$. $sum a_n =sum a_{n_k} <sum frac 1 {2^{k}} <infty$ and $sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
add a comment |
$begingroup$
We can find integers $n_1<n_2<cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n notin {n_1,n_2,cdots}$ and $a_{n_k}=q_{n_k}^{-1}$. $sum a_n =sum a_{n_k} <sum frac 1 {2^{k}} <infty$ and $sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
add a comment |
$begingroup$
We can find integers $n_1<n_2<cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n notin {n_1,n_2,cdots}$ and $a_{n_k}=q_{n_k}^{-1}$. $sum a_n =sum a_{n_k} <sum frac 1 {2^{k}} <infty$ and $sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
$endgroup$
We can find integers $n_1<n_2<cdots$ such that $q_{n_k} >2^{k}$. Let $a_n=0$ if $n notin {n_1,n_2,cdots}$ and $a_{n_k}=q_{n_k}^{-1}$. $sum a_n =sum a_{n_k} <sum frac 1 {2^{k}} <infty$ and $sum a_nq_n$ does not converge because the general term of the series does not tend to $0$.
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
edited Mar 25 at 11:50
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
answered Mar 25 at 10:29
Kavi Rama MurthyKavi Rama Murthy
75.9k53270
75.9k53270
add a comment |
add a comment |
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$begingroup$
Why not let $a_n$ be the terms of a convergent nonalternating series and render $q_n=1/a_n$? Am I missing something subtle in the language here?
$endgroup$
– Oscar Lanzi
Mar 25 at 10:07
$begingroup$
@OscarLanzi $q_n$ is given to you, you're not free to define it.
$endgroup$
– Tony S.F.
Mar 25 at 10:07
$begingroup$
You could define $a_n = 1_n/q_n$ where $1_n$ is either 0 or 1; $1_n=1$ only when $q_n>0$ and $q_n$ is "sufficiently large;" $1_n=1$ for an infinite number of indices $n$. It remains to define a precise condition for "sufficiently large."
$endgroup$
– Michael
Mar 25 at 10:29