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If the limit exists and is smaller than 1 then $limsup sqrt[n] leq lim_nrightarrowinfty |fraca_n+1a_n|$

Limit of $a_n^1/n$ is equal to $lim_ntoinfty a_n+1/a_n$Is this proof that if $a_n+1 = sqrt2 + sqrta_n$ and $a_1 = sqrt2$, then $sqrt2 leq a_n leq 2$ correct?Proofs with limit superior and limit inferior: $liminf a_n leq limsup a_n$Proof correctness of: $limsup a_n = infty implies existsa_n_k$ such that $a_n_k to infty$Prove that $limsup _nto infty(a_n+b_n)leq limsup _nto inftya_n + limsup _nto inftyb_n$Prove that $limsup a_n = sup P$ and $liminf a_n = inf P$, where $P$ is the set of limit pointsIf $fraca_n+1a_n$ converges to a real value L, then $sqrt[n]a_n$ converges to $L$.Suppose that $limsup a_n=M$ and $lim b_n= b>0$ ($bneqinfty$) as $nrightarrowinfty$,and show that $limsup a_nb_n=(limsup a_n)b$Algebra of Limit Superior, Proof Verification of $limsup (a_n + b_n) leq limsup (a_n) + limsup (b_n)$If $lim_n rightarrow infty a_n = L > 0$. Prove. $lim_n rightarrow infty sqrta_n = sqrtL$ converges

2

$begingroup$

I don't know how I can prove this Theorem. I just want to show that $limsup<1$ it does not have to be necessarily $leq lim$ in the proof Maybe this would make it easier?

My efforts:

Let $lim <q < 1$ then $|fraca_n+1a_n|leq$ for $ngeq N$.

Let $n>N$ then $|a_n| leq |a_N|q^n-N$

I want to use this for my estimate but I don't know how, in particular why can $limsup$ not be $=1$ under those conditions?

$$sqrt[n]leqsqrt[n]=q^fracn-Nnsqrt[n]=frac1q^fracNnsqrt[n]=fracsqrt[n]sqrt[n]q^N$$

Nominator converges to $1$ and denominator converges to $1$.

Can somebody help me with a hint?

Edit:

$q^fracn-Nn=qcdot q^-fracNn$

real-analysis

share|cite|improve this question

edited Mar 22 at 14:57

New2Math

asked Mar 22 at 14:22

New2MathNew2Math

16715

$endgroup$

add a comment | 

2

$begingroup$

I don't know how I can prove this Theorem. I just want to show that $limsup<1$ it does not have to be necessarily $leq lim$ in the proof Maybe this would make it easier?

My efforts:

Let $lim <q < 1$ then $|fraca_n+1a_n|leq$ for $ngeq N$.

Let $n>N$ then $|a_n| leq |a_N|q^n-N$

I want to use this for my estimate but I don't know how, in particular why can $limsup$ not be $=1$ under those conditions?

$$sqrt[n]leqsqrt[n]=q^fracn-Nnsqrt[n]=frac1q^fracNnsqrt[n]=fracsqrt[n]sqrt[n]q^N$$

Nominator converges to $1$ and denominator converges to $1$.

Can somebody help me with a hint?

Edit:

$q^fracn-Nn=qcdot q^-fracNn$

real-analysis

share|cite|improve this question

edited Mar 22 at 14:57

New2Math

asked Mar 22 at 14:22

New2MathNew2Math

16715

$endgroup$

add a comment | 

$begingroup$

I don't know how I can prove this Theorem. I just want to show that $limsup<1$ it does not have to be necessarily $leq lim$ in the proof Maybe this would make it easier?

My efforts:

Let $lim <q < 1$ then $|fraca_n+1a_n|leq$ for $ngeq N$.

Let $n>N$ then $|a_n| leq |a_N|q^n-N$

I want to use this for my estimate but I don't know how, in particular why can $limsup$ not be $=1$ under those conditions?

$$sqrt[n]leqsqrt[n]=q^fracn-Nnsqrt[n]=frac1q^fracNnsqrt[n]=fracsqrt[n]sqrt[n]q^N$$

Nominator converges to $1$ and denominator converges to $1$.

Can somebody help me with a hint?

Edit:

$q^fracn-Nn=qcdot q^-fracNn$

real-analysis

share|cite|improve this question

edited Mar 22 at 14:57

New2Math

asked Mar 22 at 14:22

New2MathNew2Math

16715

$endgroup$

share|cite|improve this question

edited Mar 22 at 14:57

New2Math

asked Mar 22 at 14:22

New2MathNew2Math

16715

share|cite|improve this question

edited Mar 22 at 14:57

New2Math

asked Mar 22 at 14:22

New2MathNew2Math

16715

asked Mar 22 at 14:22

New2MathNew2Math

16715

asked Mar 22 at 14:22

New2MathNew2Math

16715

1 Answer
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oldest

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1

$begingroup$

You don't need $q < 1$ at all. Assume that $a_n > 0$ for all $n$ and that $q > limsup dfraca_n+1a_n$.

As you observe there exists an index $N$ with the property that $$n ge N implies fraca_n+1a_n < q$$ which leads to
$$ n > N implies a_n < a_N q^n-N.$$
As long as $n > N$ it follows that
$$sqrt[n]a_n < q sqrt[n]dfraca_Nq^N. $$
It is well known that for any $a > 0$ that $sqrt[n]a to 1$, so in particular $sqrt[n]dfraca_Nq^N = 1$ because $N$ is fixed. Now calculate the limsup above:
$$ limsup_n to infty sqrt[n]a_n le limsup_n to infty qsqrt[n]dfraca_Nq^N = q lim_n to infty sqrt[n]dfraca_Nq^N = q$$
This is independent of $N$ so finally take the infimum over all $q > limsup dfraca_n+1a_n$ to find $$limsup_n to infty sqrt[n]a_n le limsup dfraca_n+1a_n.$$

share|cite|improve this answer

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

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1

$begingroup$

You don't need $q < 1$ at all. Assume that $a_n > 0$ for all $n$ and that $q > limsup dfraca_n+1a_n$.

As you observe there exists an index $N$ with the property that $$n ge N implies fraca_n+1a_n < q$$ which leads to
$$ n > N implies a_n < a_N q^n-N.$$
As long as $n > N$ it follows that
$$sqrt[n]a_n < q sqrt[n]dfraca_Nq^N. $$
It is well known that for any $a > 0$ that $sqrt[n]a to 1$, so in particular $sqrt[n]dfraca_Nq^N = 1$ because $N$ is fixed. Now calculate the limsup above:
$$ limsup_n to infty sqrt[n]a_n le limsup_n to infty qsqrt[n]dfraca_Nq^N = q lim_n to infty sqrt[n]dfraca_Nq^N = q$$
This is independent of $N$ so finally take the infimum over all $q > limsup dfraca_n+1a_n$ to find $$limsup_n to infty sqrt[n]a_n le limsup dfraca_n+1a_n.$$

share|cite|improve this answer

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

1

$begingroup$

You don't need $q < 1$ at all. Assume that $a_n > 0$ for all $n$ and that $q > limsup dfraca_n+1a_n$.

As you observe there exists an index $N$ with the property that $$n ge N implies fraca_n+1a_n < q$$ which leads to
$$ n > N implies a_n < a_N q^n-N.$$
As long as $n > N$ it follows that
$$sqrt[n]a_n < q sqrt[n]dfraca_Nq^N. $$
It is well known that for any $a > 0$ that $sqrt[n]a to 1$, so in particular $sqrt[n]dfraca_Nq^N = 1$ because $N$ is fixed. Now calculate the limsup above:
$$ limsup_n to infty sqrt[n]a_n le limsup_n to infty qsqrt[n]dfraca_Nq^N = q lim_n to infty sqrt[n]dfraca_Nq^N = q$$
This is independent of $N$ so finally take the infimum over all $q > limsup dfraca_n+1a_n$ to find $$limsup_n to infty sqrt[n]a_n le limsup dfraca_n+1a_n.$$

share|cite|improve this answer

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

$begingroup$

You don't need $q < 1$ at all. Assume that $a_n > 0$ for all $n$ and that $q > limsup dfraca_n+1a_n$.

As you observe there exists an index $N$ with the property that $$n ge N implies fraca_n+1a_n < q$$ which leads to
$$ n > N implies a_n < a_N q^n-N.$$
As long as $n > N$ it follows that
$$sqrt[n]a_n < q sqrt[n]dfraca_Nq^N. $$
It is well known that for any $a > 0$ that $sqrt[n]a to 1$, so in particular $sqrt[n]dfraca_Nq^N = 1$ because $N$ is fixed. Now calculate the limsup above:
$$ limsup_n to infty sqrt[n]a_n le limsup_n to infty qsqrt[n]dfraca_Nq^N = q lim_n to infty sqrt[n]dfraca_Nq^N = q$$
This is independent of $N$ so finally take the infimum over all $q > limsup dfraca_n+1a_n$ to find $$limsup_n to infty sqrt[n]a_n le limsup dfraca_n+1a_n.$$

share|cite|improve this answer

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370

answered Mar 22 at 15:28

Umberto P.Umberto P.

40.3k13370