Mixing
Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges(2)...
$begingroup$
This question already has an answer here:
Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
2 answers
Suppose that ${a_n}$ be a sequence of real numbers.let ${n_{k}}$ be an increasing sequence of positive integers, and let $A_{k} = a_{n_{k}} + a_{n_{k}+1}+ ... +a_{n_{k +1}-1}$, for each $k in mathbb{N}$. Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges. Give an example to show that the converse is not true.
Can anyone explain for me in the term $a_{n_{k+1} -1}$ in the definition of $A_k$ we take a decreasing index (i.e why we subtract 1 and not add 1 to $n_{k+1}$ as we continued adding 1 to $n_{k}$ ) or this is a typo?
Also can anyone give me a numeric example for the direction we want to prove ?
Also I received a hint about the proof that I should use Cauchy criterion, could anyone provide me with a little bit more details in this direction?
Those are related links:
1- Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges.
2-Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
Thank you!
calculus sequences-and-series cauchy-sequences
asked Jan 15 at 7:48
hopefullyhopefully
240114
$endgroup$
marked as duplicate by RRL, José Carlos Santos calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 15 at 8:26
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 |
$begingroup$
This question already has an answer here:
Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
2 answers
Suppose that ${a_n}$ be a sequence of real numbers.let ${n_{k}}$ be an increasing sequence of positive integers, and let $A_{k} = a_{n_{k}} + a_{n_{k}+1}+ ... +a_{n_{k +1}-1}$, for each $k in mathbb{N}$. Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges. Give an example to show that the converse is not true.
Can anyone explain for me in the term $a_{n_{k+1} -1}$ in the definition of $A_k$ we take a decreasing index (i.e why we subtract 1 and not add 1 to $n_{k+1}$ as we continued adding 1 to $n_{k}$ ) or this is a typo?
Also can anyone give me a numeric example for the direction we want to prove ?
Also I received a hint about the proof that I should use Cauchy criterion, could anyone provide me with a little bit more details in this direction?
Those are related links:
1- Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges.
2-Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
Thank you!
calculus sequences-and-series cauchy-sequences
asked Jan 15 at 7:48
hopefullyhopefully
240114
$endgroup$
marked as duplicate by RRL, José Carlos Santos calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 15 at 8:26
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 |
$begingroup$
This question already has an answer here:
Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
2 answers
Suppose that ${a_n}$ be a sequence of real numbers.let ${n_{k}}$ be an increasing sequence of positive integers, and let $A_{k} = a_{n_{k}} + a_{n_{k}+1}+ ... +a_{n_{k +1}-1}$, for each $k in mathbb{N}$. Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges. Give an example to show that the converse is not true.
Can anyone explain for me in the term $a_{n_{k+1} -1}$ in the definition of $A_k$ we take a decreasing index (i.e why we subtract 1 and not add 1 to $n_{k+1}$ as we continued adding 1 to $n_{k}$ ) or this is a typo?
Also can anyone give me a numeric example for the direction we want to prove ?
Also I received a hint about the proof that I should use Cauchy criterion, could anyone provide me with a little bit more details in this direction?
Those are related links:
1- Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges.
2-Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
Thank you!
calculus sequences-and-series cauchy-sequences
asked Jan 15 at 7:48
hopefullyhopefully
240114
$endgroup$
This question already has an answer here:
Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
2 answers
Suppose that ${a_n}$ be a sequence of real numbers.let ${n_{k}}$ be an increasing sequence of positive integers, and let $A_{k} = a_{n_{k}} + a_{n_{k}+1}+ ... +a_{n_{k +1}-1}$, for each $k in mathbb{N}$. Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges. Give an example to show that the converse is not true.
Can anyone explain for me in the term $a_{n_{k+1} -1}$ in the definition of $A_k$ we take a decreasing index (i.e why we subtract 1 and not add 1 to $n_{k+1}$ as we continued adding 1 to $n_{k}$ ) or this is a typo?
Also can anyone give me a numeric example for the direction we want to prove ?
Also I received a hint about the proof that I should use Cauchy criterion, could anyone provide me with a little bit more details in this direction?
Those are related links:
1- Prove that if $sum_{n = 1}^{infty} a_{n}$ converges, then $sum_{k = 1}^{infty} A_{k}$ converges.
2-Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
Thank you!
This question already has an answer here:
Prove that if $sum_{n=1}^infty a_n$ converges then $sum_{k=1}^infty A_k$ converges
2 answers
calculus sequences-and-series cauchy-sequences
calculus sequences-and-series cauchy-sequences
asked Jan 15 at 7:48
hopefullyhopefully
240114
asked Jan 15 at 7:48
hopefullyhopefully
240114
asked Jan 15 at 7:48
hopefullyhopefully
240114
asked Jan 15 at 7:48
hopefullyhopefully
240114
asked Jan 15 at 7:48
hopefullyhopefully
240114
240114
marked as duplicate by RRL, José Carlos Santos calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 15 at 8:26
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 RRL, José Carlos Santos calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 15 at 8:26
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
1
active
oldest
votes
$begingroup$
It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where ${S_n}$ is the partial sum sequence of $sum a_n$. Hence $A_1+A_2+cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since ${S_n}$ has a finite limit it is now clear that $sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+cdots$ . This series is not convergent but $A_1=0,A_2=0,cdots$. Hence $sum A_k$ is convergent even though $sum a_k$ is not.
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
$endgroup$
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where ${S_n}$ is the partial sum sequence of $sum a_n$. Hence $A_1+A_2+cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since ${S_n}$ has a finite limit it is now clear that $sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+cdots$ . This series is not convergent but $A_1=0,A_2=0,cdots$. Hence $sum A_k$ is convergent even though $sum a_k$ is not.
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
$endgroup$
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
add a comment |
$begingroup$
It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where ${S_n}$ is the partial sum sequence of $sum a_n$. Hence $A_1+A_2+cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since ${S_n}$ has a finite limit it is now clear that $sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+cdots$ . This series is not convergent but $A_1=0,A_2=0,cdots$. Hence $sum A_k$ is convergent even though $sum a_k$ is not.
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
$endgroup$
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
add a comment |
$begingroup$
It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where ${S_n}$ is the partial sum sequence of $sum a_n$. Hence $A_1+A_2+cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since ${S_n}$ has a finite limit it is now clear that $sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+cdots$ . This series is not convergent but $A_1=0,A_2=0,cdots$. Hence $sum A_k$ is convergent even though $sum a_k$ is not.
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
$endgroup$
It is important not to include $a_{n_{k+1}}$ in $A_k$ to avoid this term getting repeated in $A_k$ as well as $A_{k+1}$. To prove the convergence of $sum A_k$ note that we can write $A_k$ as $S_{n_{k+1}} -S_{n_k}$ where ${S_n}$ is the partial sum sequence of $sum a_n$. Hence $A_1+A_2+cdots+A_N= (S_{n_2} -S_{n_1})+(S_{n_{3}}- S_{n_2})+cdots+S_{n_{N+1}} -S_{n_N}=S_{n_{N+1}}-S_{n_1}$. Since ${S_n}$ has a finite limit it is now clear that $sum A_k$ converges. Example: take $n_k=2k$ as see what happens. To prove that the converse is not true take $n_k=2k$. Consider the series $1+(-1)+1+(-1)+cdots$ . This series is not convergent but $A_1=0,A_2=0,cdots$. Hence $sum A_k$ is convergent even though $sum a_k$ is not.
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
answered Jan 15 at 7:56
Kavi Rama MurthyKavi Rama Murthy
60.9k42161
60.9k42161
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
add a comment |
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
in the numeric example that clarify the direction that we required to prove what series should I consider?
$endgroup$
– hopefully
Jan 15 at 8:09
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
@hopefully I what I suggested was to write down the sequence ${A_k}$ in terms of $a_n$'s when $n_k=2k$. This should give you an idea as to why $sum A_k$ converges whenever $sum A_k$ converges.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 8:15
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
$begingroup$
what about this sentence mentioned in one of the answers in one of the links above "(If $n_1neq 1$, the you're only missing finitely many initial terms, so you're still good)"...... could you explain it for me please?
$endgroup$
– hopefully
Jan 15 at 8:38
add a comment |
