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Given that $left(frac{a^2}{2}a_n-ab_nright)_n$ converges for all $ain(0,1]$, show that $(a_n)_n,(b_n)_n$ are convergent

$sum_{i=1}^{infty}a_n*b_n $ converges for all $lim_{n rightarrow infty}b_n = 1$, show that $a$ converges absolutelyShow that $lim_{n to infty} a_n = infty$ if and only if $lim_{n to infty} frac{1}{a_n} = 0$Prove that if $a_n>0$ and $sum a_n$ converges then $sum (frac {b_n}{a_n})$ 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$left{a_n, b_n right}$. Proving convergenceProb. 8, Chap. 3 in Baby Rudin: If $sum a_n$ converges and $left{b_nright}$ 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

2

$begingroup$

We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac{1}{2}a_n-b_nright)_n$ is convergent and $left(frac{a^2}{2}a_n-ab_nright)_n$ 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

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edited Mar 9 at 20:52

rtybase

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asked Mar 9 at 17:19

GaboruGaboru

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2

$begingroup$

We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac{1}{2}a_n-b_nright)_n$ is convergent and $left(frac{a^2}{2}a_n-ab_nright)_n$ 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

share|cite|improve this question

edited Mar 9 at 20:52

rtybase

11.4k31533

asked Mar 9 at 17:19

GaboruGaboru

4237

$endgroup$

add a comment | 

$begingroup$

We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $left(frac{1}{2}a_n-b_nright)_n$ is convergent and $left(frac{a^2}{2}a_n-ab_nright)_n$ 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

share|cite|improve this question

edited Mar 9 at 20:52

rtybase

11.4k31533

asked Mar 9 at 17:19

GaboruGaboru

4237

$endgroup$

share|cite|improve this question

edited Mar 9 at 20:52

rtybase

11.4k31533

asked Mar 9 at 17:19

GaboruGaboru

4237

share|cite|improve this question

edited Mar 9 at 20:52

rtybase

11.4k31533

asked Mar 9 at 17:19

GaboruGaboru

4237

edited Mar 9 at 20:52

rtybase

11.4k31533

edited Mar 9 at 20:52

rtybase

11.4k31533

asked Mar 9 at 17:19

GaboruGaboru

4237

asked Mar 9 at 17:19

GaboruGaboru

4237

2 Answers
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1

$begingroup$

Since $frac{a^2}{2}a_n-ab_n$ converges for $a in (0,1]$, also $frac{1}{a}cdot (frac{a^2}{2}a_n-ab_n) = frac{a}{2}a_n-b_n$ converges. Hence $(frac{a}{2}a_n-b_n) - (frac{1}{2}a_n-b_n) = frac{a-1}{2}a_n$ converges. We conclude that $frac{2}{a-1} cdot frac{a-1}{2}a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac{1}{2}cdot a_n - (frac{1}{2}a_n-b_n) = b_n$ converges.

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edited Mar 9 at 19:34

answered Mar 9 at 17:37

Paul FrostPaul Frost

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0

$begingroup$

Let $f_n(a):=tfrac{a^2}{2}a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext{ and }qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$

share|cite|improve this answer

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

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1

$begingroup$

Since $frac{a^2}{2}a_n-ab_n$ converges for $a in (0,1]$, also $frac{1}{a}cdot (frac{a^2}{2}a_n-ab_n) = frac{a}{2}a_n-b_n$ converges. Hence $(frac{a}{2}a_n-b_n) - (frac{1}{2}a_n-b_n) = frac{a-1}{2}a_n$ converges. We conclude that $frac{2}{a-1} cdot frac{a-1}{2}a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac{1}{2}cdot a_n - (frac{1}{2}a_n-b_n) = b_n$ converges.

share|cite|improve this answer

edited Mar 9 at 19:34

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

$endgroup$

add a comment | 

1

$begingroup$

Since $frac{a^2}{2}a_n-ab_n$ converges for $a in (0,1]$, also $frac{1}{a}cdot (frac{a^2}{2}a_n-ab_n) = frac{a}{2}a_n-b_n$ converges. Hence $(frac{a}{2}a_n-b_n) - (frac{1}{2}a_n-b_n) = frac{a-1}{2}a_n$ converges. We conclude that $frac{2}{a-1} cdot frac{a-1}{2}a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac{1}{2}cdot a_n - (frac{1}{2}a_n-b_n) = b_n$ converges.

share|cite|improve this answer

edited Mar 9 at 19:34

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

$endgroup$

add a comment | 

$begingroup$

Since $frac{a^2}{2}a_n-ab_n$ converges for $a in (0,1]$, also $frac{1}{a}cdot (frac{a^2}{2}a_n-ab_n) = frac{a}{2}a_n-b_n$ converges. Hence $(frac{a}{2}a_n-b_n) - (frac{1}{2}a_n-b_n) = frac{a-1}{2}a_n$ converges. We conclude that $frac{2}{a-1} cdot frac{a-1}{2}a_n = a_n$ converges (note $a-1 ne 0$ for $a < 1$). Hence also $frac{1}{2}cdot a_n - (frac{1}{2}a_n-b_n) = b_n$ converges.

share|cite|improve this answer

edited Mar 9 at 19:34

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

$endgroup$

share|cite|improve this answer

edited Mar 9 at 19:34

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

answered Mar 9 at 17:37

Paul FrostPaul Frost

11.6k3934

0

$begingroup$

Let $f_n(a):=tfrac{a^2}{2}a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext{ and }qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$

share|cite|improve this answer

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

$endgroup$

add a comment | 

0

$begingroup$

Let $f_n(a):=tfrac{a^2}{2}a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext{ and }qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$

share|cite|improve this answer

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

$endgroup$

add a comment | 

$begingroup$

Let $f_n(a):=tfrac{a^2}{2}a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $ain(0,1]$, so do
$$(4f_n(1)-8f_n(tfrac12))_n=(a_n)_n
qquadtext{ and }qquad
(f_n(1)-4f_n(tfrac12))_n=(b_n)_n.$$

share|cite|improve this answer

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

$endgroup$

share|cite|improve this answer

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

answered Mar 9 at 19:53

ServaesServaes

28.2k34099

answered Mar 9 at 19:53

ServaesServaes

28.2k34099