If ${a_n}$ is bounded and non-decreasing, prove that $liminf b_n = 0$, $b_n = n(a_{n+1} - a_{n})$












2












$begingroup$


Let ${a_n}$ be a bounded and non-decreasing sequence of reals, and $b_n = n(a_{n+1} - a_{n})$.



(a) Show that $liminf b_n = 0$



(b) Give an example of a sequence ${a_n}$ such that ${b_n}$ diverges.



I know that $a_{n+1} geq a_{n}, forall n$,



Hence, $a_{n+1} - a_{n} geq 0, forall n$.



Also, we have that $exists B >0$ s.t $|a_n| leq B, forall n$.



Hence, $b_n geq 0$ and $|b_n| leq n2B$



If ${a_n}$ is bounded, then ${a_n}$ converges by Bolzano Weierstrass Theorem, since ${a_n} subseteq [-B,B]$



Now, since ${a_n}$ converges, there exists a subsequence ${a_{n_k}}$ of bounded variation, hence $forall epsilon >0, exists N >0, forall n_kgeq N$,



$|a_{n_{k+1}} - a_{n_k}| < epsilon$



But I can't proof that the $liminf b_n =0$.






real-analysis sequences-and-series cauchy-sequences limsup-and-liminf







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Assume that $liminf b_n > 0$.
    $endgroup$
    – Martin R
    Jan 11 at 14:21








  • 1




    $begingroup$
    ... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
    $endgroup$
    – Song
    Jan 11 at 14:22
















2












$begingroup$


Let ${a_n}$ be a bounded and non-decreasing sequence of reals, and $b_n = n(a_{n+1} - a_{n})$.



(a) Show that $liminf b_n = 0$



(b) Give an example of a sequence ${a_n}$ such that ${b_n}$ diverges.



I know that $a_{n+1} geq a_{n}, forall n$,



Hence, $a_{n+1} - a_{n} geq 0, forall n$.



Also, we have that $exists B >0$ s.t $|a_n| leq B, forall n$.



Hence, $b_n geq 0$ and $|b_n| leq n2B$



If ${a_n}$ is bounded, then ${a_n}$ converges by Bolzano Weierstrass Theorem, since ${a_n} subseteq [-B,B]$



Now, since ${a_n}$ converges, there exists a subsequence ${a_{n_k}}$ of bounded variation, hence $forall epsilon >0, exists N >0, forall n_kgeq N$,



$|a_{n_{k+1}} - a_{n_k}| < epsilon$



But I can't proof that the $liminf b_n =0$.






real-analysis sequences-and-series cauchy-sequences limsup-and-liminf







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: Assume that $liminf b_n > 0$.
    $endgroup$
    – Martin R
    Jan 11 at 14:21








  • 1




    $begingroup$
    ... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
    $endgroup$
    – Song
    Jan 11 at 14:22














2












2








2





$begingroup$


Let ${a_n}$ be a bounded and non-decreasing sequence of reals, and $b_n = n(a_{n+1} - a_{n})$.



(a) Show that $liminf b_n = 0$



(b) Give an example of a sequence ${a_n}$ such that ${b_n}$ diverges.



I know that $a_{n+1} geq a_{n}, forall n$,



Hence, $a_{n+1} - a_{n} geq 0, forall n$.



Also, we have that $exists B >0$ s.t $|a_n| leq B, forall n$.



Hence, $b_n geq 0$ and $|b_n| leq n2B$



If ${a_n}$ is bounded, then ${a_n}$ converges by Bolzano Weierstrass Theorem, since ${a_n} subseteq [-B,B]$



Now, since ${a_n}$ converges, there exists a subsequence ${a_{n_k}}$ of bounded variation, hence $forall epsilon >0, exists N >0, forall n_kgeq N$,



$|a_{n_{k+1}} - a_{n_k}| < epsilon$



But I can't proof that the $liminf b_n =0$.






real-analysis sequences-and-series cauchy-sequences limsup-and-liminf







share|cite|improve this question









$endgroup$




Let ${a_n}$ be a bounded and non-decreasing sequence of reals, and $b_n = n(a_{n+1} - a_{n})$.



(a) Show that $liminf b_n = 0$



(b) Give an example of a sequence ${a_n}$ such that ${b_n}$ diverges.



I know that $a_{n+1} geq a_{n}, forall n$,



Hence, $a_{n+1} - a_{n} geq 0, forall n$.



Also, we have that $exists B >0$ s.t $|a_n| leq B, forall n$.



Hence, $b_n geq 0$ and $|b_n| leq n2B$



If ${a_n}$ is bounded, then ${a_n}$ converges by Bolzano Weierstrass Theorem, since ${a_n} subseteq [-B,B]$



Now, since ${a_n}$ converges, there exists a subsequence ${a_{n_k}}$ of bounded variation, hence $forall epsilon >0, exists N >0, forall n_kgeq N$,



$|a_{n_{k+1}} - a_{n_k}| < epsilon$



But I can't proof that the $liminf b_n =0$.






real-analysis sequences-and-series cauchy-sequences limsup-and-liminf




real-analysis sequences-and-series cauchy-sequences limsup-and-liminf






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 14:16









Richard ClareRichard Clare

1,066314




1,066314








  • 1




    $begingroup$
    Hint: Assume that $liminf b_n > 0$.
    $endgroup$
    – Martin R
    Jan 11 at 14:21








  • 1




    $begingroup$
    ... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
    $endgroup$
    – Song
    Jan 11 at 14:22














  • 1




    $begingroup$
    Hint: Assume that $liminf b_n > 0$.
    $endgroup$
    – Martin R
    Jan 11 at 14:21








  • 1




    $begingroup$
    ... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
    $endgroup$
    – Song
    Jan 11 at 14:22








1




1




$begingroup$
Hint: Assume that $liminf b_n > 0$.
$endgroup$
– Martin R
Jan 11 at 14:21






$begingroup$
Hint: Assume that $liminf b_n > 0$.
$endgroup$
– Martin R
Jan 11 at 14:21






1




1




$begingroup$
... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
$endgroup$
– Song
Jan 11 at 14:22




$begingroup$
... Then there exists $epsilon>0$ and $Ninmathbb{N}$ such that $b_nge epsilon$ for all $nge N$ ...
$endgroup$
– Song
Jan 11 at 14:22










1 Answer
1






active

oldest

votes


















3












$begingroup$

Assume that $c = liminf b_n > 0$. Then there is a $N in Bbb N$ such that
$$
n(a_{n+1} - a_{n}) = b_n > frac c2 > 0
implies a_{n+1} - a_n > frac{c}{2n}
$$
for all $n ge N$. Summing the last inequality gives
$$
a_{n} = a_N + sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + frac c2 sum_{j=N}^{n-1} frac 1j
$$
for $n ge N$. Since the harmonic series diverges, it follows that
$a_n to infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:45










  • $begingroup$
    @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
    $endgroup$
    – Martin R
    Jan 11 at 14:50












  • $begingroup$
    @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
    $endgroup$
    – Martin R
    Jan 11 at 14:53










  • $begingroup$
    Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:54












  • $begingroup$
    I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
    $endgroup$
    – Richard Clare
    Jan 11 at 14:57













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Assume that $c = liminf b_n > 0$. Then there is a $N in Bbb N$ such that
$$
n(a_{n+1} - a_{n}) = b_n > frac c2 > 0
implies a_{n+1} - a_n > frac{c}{2n}
$$
for all $n ge N$. Summing the last inequality gives
$$
a_{n} = a_N + sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + frac c2 sum_{j=N}^{n-1} frac 1j
$$
for $n ge N$. Since the harmonic series diverges, it follows that
$a_n to infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:45










  • $begingroup$
    @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
    $endgroup$
    – Martin R
    Jan 11 at 14:50












  • $begingroup$
    @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
    $endgroup$
    – Martin R
    Jan 11 at 14:53










  • $begingroup$
    Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:54












  • $begingroup$
    I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
    $endgroup$
    – Richard Clare
    Jan 11 at 14:57


















3












$begingroup$

Assume that $c = liminf b_n > 0$. Then there is a $N in Bbb N$ such that
$$
n(a_{n+1} - a_{n}) = b_n > frac c2 > 0
implies a_{n+1} - a_n > frac{c}{2n}
$$
for all $n ge N$. Summing the last inequality gives
$$
a_{n} = a_N + sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + frac c2 sum_{j=N}^{n-1} frac 1j
$$
for $n ge N$. Since the harmonic series diverges, it follows that
$a_n to infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:45










  • $begingroup$
    @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
    $endgroup$
    – Martin R
    Jan 11 at 14:50












  • $begingroup$
    @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
    $endgroup$
    – Martin R
    Jan 11 at 14:53










  • $begingroup$
    Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:54












  • $begingroup$
    I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
    $endgroup$
    – Richard Clare
    Jan 11 at 14:57
















3












3








3





$begingroup$

Assume that $c = liminf b_n > 0$. Then there is a $N in Bbb N$ such that
$$
n(a_{n+1} - a_{n}) = b_n > frac c2 > 0
implies a_{n+1} - a_n > frac{c}{2n}
$$
for all $n ge N$. Summing the last inequality gives
$$
a_{n} = a_N + sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + frac c2 sum_{j=N}^{n-1} frac 1j
$$
for $n ge N$. Since the harmonic series diverges, it follows that
$a_n to infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.






share|cite|improve this answer











$endgroup$



Assume that $c = liminf b_n > 0$. Then there is a $N in Bbb N$ such that
$$
n(a_{n+1} - a_{n}) = b_n > frac c2 > 0
implies a_{n+1} - a_n > frac{c}{2n}
$$
for all $n ge N$. Summing the last inequality gives
$$
a_{n} = a_N + sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + frac c2 sum_{j=N}^{n-1} frac 1j
$$
for $n ge N$. Since the harmonic series diverges, it follows that
$a_n to infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 15:01

























answered Jan 11 at 14:25









Martin RMartin R

27.9k33255




27.9k33255












  • $begingroup$
    We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:45










  • $begingroup$
    @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
    $endgroup$
    – Martin R
    Jan 11 at 14:50












  • $begingroup$
    @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
    $endgroup$
    – Martin R
    Jan 11 at 14:53










  • $begingroup$
    Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:54












  • $begingroup$
    I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
    $endgroup$
    – Richard Clare
    Jan 11 at 14:57




















  • $begingroup$
    We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:45










  • $begingroup$
    @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
    $endgroup$
    – Martin R
    Jan 11 at 14:50












  • $begingroup$
    @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
    $endgroup$
    – Martin R
    Jan 11 at 14:53










  • $begingroup$
    Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
    $endgroup$
    – Richard Clare
    Jan 11 at 14:54












  • $begingroup$
    I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
    $endgroup$
    – Richard Clare
    Jan 11 at 14:57


















$begingroup$
We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
$endgroup$
– Richard Clare
Jan 11 at 14:45




$begingroup$
We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ?
$endgroup$
– Richard Clare
Jan 11 at 14:45












$begingroup$
@RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
$endgroup$
– Martin R
Jan 11 at 14:50






$begingroup$
@RichardClare: Summing the above relationship gives $a_{n+1} > a_N + frac c2 sum_{j=N}^n frac 1j to infty$ for $n to infty$, contrary to the assumption that $(a_n)$ is bounded.
$endgroup$
– Martin R
Jan 11 at 14:50














$begingroup$
@RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
$endgroup$
– Martin R
Jan 11 at 14:53




$begingroup$
@RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n ge N}$ implies the divergence of the “entire” sequence $(a_n)$.
$endgroup$
– Martin R
Jan 11 at 14:53












$begingroup$
Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
$endgroup$
– Richard Clare
Jan 11 at 14:54






$begingroup$
Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ?
$endgroup$
– Richard Clare
Jan 11 at 14:54














$begingroup$
I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
$endgroup$
– Richard Clare
Jan 11 at 14:57






$begingroup$
I got this: $sum_{n = N}^{infty}(a_{n+1} - a_n) > frac c2 sum_{j=N}^n frac 1j to infty$
$endgroup$
– Richard Clare
Jan 11 at 14:57




















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