Given that $left(fraca^22a_n-ab_nright)_n$ converges for all $ain(0,1]$, show that $(a_n)_n,(b_n)_n$ are convergent$sum_i=1^inftya_n*b_n $ converges for all $lim_n rightarrow inftyb_n = 1$, show that $a$ converges absolutelyShow that $lim_n to infty a_n = infty$ if and only if $lim_n to infty frac1a_n = 0$Prove that if $a_n>0$ and $sum a_n$ converges then $sum (frac b_na_n)$ convergesIf $a_nb_n$ is absolutely convergent for every $b_n$ that converges to 0, then $a_n$ is absolutely convergent.Sequence defined by max$lefta_n, b_n right$. Proving convergenceProb. 8, Chap. 3 in Baby Rudin: If $sum a_n$ converges and $leftb_nright$ is monotonic and bounded, then $sum a_n b_n$ converges.If $a_n$ and $a_nb_n$ are convergent sequences, then $b_n$ converges.$a_n$ is convergent, $b_n$ bounded, prove $sum a_n b_n$ convergesGiven $a_1 =1, a_n+1=a_n+e^-a_n$. Prove that $b_n:=a_n-ln(n)$ converges$limlimits_nto infty int_0^1 |f(x)-a_nx-b_n| dx=0$ implies $(a_n)_n,(b_n)_n$ are convergent
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Given that $left(fraca^22a_n-ab_nright)_n$ converges for all $ain(0,1]$, show that $(a_n)_n,(b_n)_n$ are convergent
$sum_i=1^inftya_n*b_n $ converges for all $lim_n rightarrow inftyb_n = 1$, show that $a$ converges absolutelyShow that $lim_n to infty a_n = infty$ if and only if $lim_n to infty frac1a_n = 0$Prove that if $a_n>0$ and $sum a_n$ converges then $sum (frac b_na_n)$ convergesIf $a_nb_n$ is absolutely convergent for every $b_n$ that converges to 0, then $a_n$ is absolutely convergent.Sequence defined by max$lefta_n, b_n right$. Proving convergenceProb. 8, Chap. 3 in Baby Rudin: If $sum a_n$ converges and $leftb_nright$ is monotonic and bounded, then $sum a_n b_n$ converges.If $a_n$ and $a_nb_n$ are convergent sequences, then $b_n$ converges.$a_n$ is convergent, $b_n$ bounded, prove $sum a_n b_n$ convergesGiven $a_1 =1, a_n+1=a_n+e^-a_n$. Prove that $b_n:=a_n-ln(n)$ converges$limlimits_nto infty int_0^1 |f(x)-a_nx-b_n| dx=0$ implies $(a_n)_n,(b_n)_n$ are convergent
$begingroup$
We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac12a_n-b_nright)_n$ is convergent and $left(fraca^22a_n-ab_nright)_n$ converges for all $ain (0,1]$. Can somebody help me how to prove that $(a_n)_n,(b_n)_n$ are convergent? Please
sequences-and-series algebra-precalculus limits convergence
edited 2 days ago
rtybase
11.4k31533
asked 2 days ago
GaboruGaboru
3897
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add a comment |
$begingroup$
We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac12a_n-b_nright)_n$ is convergent and $left(fraca^22a_n-ab_nright)_n$ converges for all $ain (0,1]$. Can somebody help me how to prove that $(a_n)_n,(b_n)_n$ are convergent? Please
sequences-and-series algebra-precalculus limits convergence
edited 2 days ago
rtybase
11.4k31533
asked 2 days ago
GaboruGaboru
3897
$endgroup$
add a comment |
$begingroup$
We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac12a_n-b_nright)_n$ is convergent and $left(fraca^22a_n-ab_nright)_n$ converges for all $ain (0,1]$. Can somebody help me how to prove that $(a_n)_n,(b_n)_n$ are convergent? Please
sequences-and-series algebra-precalculus limits convergence
edited 2 days ago
rtybase
11.4k31533
asked 2 days ago
GaboruGaboru
3897
$endgroup$
We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac12a_n-b_nright)_n$ is convergent and $left(fraca^22a_n-ab_nright)_n$ converges for all $ain (0,1]$. Can somebody help me how to prove that $(a_n)_n,(b_n)_n$ are convergent? Please
sequences-and-series algebra-precalculus limits convergence
sequences-and-series algebra-precalculus limits convergence
edited 2 days ago
rtybase
11.4k31533
asked 2 days ago
GaboruGaboru
3897
edited 2 days ago
rtybase
11.4k31533
asked 2 days ago
GaboruGaboru
3897
edited 2 days ago
rtybase
11.4k31533
edited 2 days ago
rtybase
11.4k31533
edited 2 days ago
rtybase
11.4k31533
11.4k31533
asked 2 days ago
GaboruGaboru
3897
asked 2 days ago
GaboruGaboru
3897
asked 2 days ago
GaboruGaboru
3897
3897
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add a comment |
2 Answers
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$begingroup$
Since $fraca^22a_n-ab_n$ converges for $a in (0,1]$, also $frac1acdot (fraca^22a_n-ab_n) = fraca2a_n-b_n$ converges. Hence $(fraca2a_n-b_n) - (frac12a_n-b_n) = fraca-12a_n$ converges. We conclude that $frac2a-1 cdot fraca-12a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac12cdot a_n - (frac12a_n-b_n) = b_n$ converges.
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
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$begingroup$
Let $f_n(a):=tfraca^22a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext and qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$
answered 2 days ago
ServaesServaes
28.1k34099
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2 Answers
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$begingroup$
Since $fraca^22a_n-ab_n$ converges for $a in (0,1]$, also $frac1acdot (fraca^22a_n-ab_n) = fraca2a_n-b_n$ converges. Hence $(fraca2a_n-b_n) - (frac12a_n-b_n) = fraca-12a_n$ converges. We conclude that $frac2a-1 cdot fraca-12a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac12cdot a_n - (frac12a_n-b_n) = b_n$ converges.
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
$endgroup$
add a comment |
$begingroup$
Since $fraca^22a_n-ab_n$ converges for $a in (0,1]$, also $frac1acdot (fraca^22a_n-ab_n) = fraca2a_n-b_n$ converges. Hence $(fraca2a_n-b_n) - (frac12a_n-b_n) = fraca-12a_n$ converges. We conclude that $frac2a-1 cdot fraca-12a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac12cdot a_n - (frac12a_n-b_n) = b_n$ converges.
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
$endgroup$
add a comment |
$begingroup$
Since $fraca^22a_n-ab_n$ converges for $a in (0,1]$, also $frac1acdot (fraca^22a_n-ab_n) = fraca2a_n-b_n$ converges. Hence $(fraca2a_n-b_n) - (frac12a_n-b_n) = fraca-12a_n$ converges. We conclude that $frac2a-1 cdot fraca-12a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac12cdot a_n - (frac12a_n-b_n) = b_n$ converges.
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
$endgroup$
Since $fraca^22a_n-ab_n$ converges for $a in (0,1]$, also $frac1acdot (fraca^22a_n-ab_n) = fraca2a_n-b_n$ converges. Hence $(fraca2a_n-b_n) - (frac12a_n-b_n) = fraca-12a_n$ converges. We conclude that $frac2a-1 cdot fraca-12a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac12cdot a_n - (frac12a_n-b_n) = b_n$ converges.
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
edited 2 days ago
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
answered 2 days ago
Paul FrostPaul Frost
11.6k3934
11.6k3934
add a comment |
add a comment |
$begingroup$
Let $f_n(a):=tfraca^22a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext and qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$
answered 2 days ago
ServaesServaes
28.1k34099
$endgroup$
add a comment |
$begingroup$
Let $f_n(a):=tfraca^22a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext and qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$
answered 2 days ago
ServaesServaes
28.1k34099
$endgroup$
add a comment |
$begingroup$
Let $f_n(a):=tfraca^22a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext and qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$
answered 2 days ago
ServaesServaes
28.1k34099
$endgroup$
Let $f_n(a):=tfraca^22a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext and qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$
answered 2 days ago
ServaesServaes
28.1k34099
answered 2 days ago
ServaesServaes
28.1k34099
answered 2 days ago
ServaesServaes
28.1k34099
answered 2 days ago
ServaesServaes
28.1k34099
28.1k34099
add a comment |
add a comment |
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